The user wants me to solve a physics problem involving a sequence of operations on a circuit with two capacitors and two voltage sources. I will analyze the circuit step-by-step.

Step-by-step Derivation:

1. Step 1: Charging Capacitor $C_1$

   • The switch $S_1$ is pressed, connecting the capacitor $C_1$ (capacitance $C$) to the voltage source $2V_0$.
   • The capacitor is fully charged, so the voltage across it becomes $2V_0$.
   • The charge on the capacitor is given by $Q=CV$. Let's denote the charge on the upper plate of $C_1$ as $q_1$.
   • $q_1 = C \times (2V_0) = 2CV_0$.
   • The lower plate has a charge of $-2CV_0$. The capacitor $C_2$ is initially uncharged, so $q_2=0$.
   • After charging, $S_1$ is released.

2. Step 2: Charging Capacitor $C_2$

   • The switch $S_2$ is pressed. This connects the charged capacitor $C_1$ into a new circuit with capacitor $C_2$ (capacitance $C$) and voltage source $V_0$.
   • Let's analyze the connections:
     • The upper plates of $C_1$ and $C_2$ are connected together.
     • The lower plate of $C_1$ is connected to the positive terminal of the $V_0$ battery.
     • The lower plate of $C_2$ is connected to the negative terminal of the $V_0$ battery.
   • Let's find the charges on the capacitors in the steady state. Let the charges on the upper plates be $q_1'$ and $q_2'$.
   • The upper plates of $C_1$ and $C_2$ along with the connecting wire form an isolated system. The total charge on this system is conserved.
   • Initial charge on the isolated system = (charge on upper plate of $C_1$) + (charge on upper plate of $C_2$) = $2CV_0 + 0 = 2CV_0$.
   • Final charge on the isolated system = $q_1' + q_2'$.
   • By charge conservation: $q_1' + q_2' = 2CV_0 \quad (1)$
   • Now, let's apply Kirchhoff's voltage law (KVL) to the loop formed by $C_1$, $C_2$, and $V_0$. Let's define the potential of the negative terminal of $V_0$ as 0. Then the potential of the positive terminal is $V_0$. The potential of the lower plate of $C_2$ is $0$, and the potential of the lower plate of $C_1$ is $V_0$. Let the common potential of the connected upper plates be $V_U$.
   • The potential difference across $C_1$ is $V_{C1} = V_U - V_0$. So, $q_1' = C(V_U - V_0)$.
   • The potential difference across $C_2$ is $V_{C2} = V_U - 0 = V_U$. So, $q_2' = CV_U$.
   • Substitute these into the charge conservation equation (1): $C(V_U - V_0) + CV_U = 2CV_0$ $2CV_U - CV_0 = 2CV_0$ $2CV_U = 3CV_0 \implies V_U = \frac{3}{2}V_0$
   • Now we can find the charges $q_1'$ and $q_2'$: $q_1' = C\left(\frac{3}{2}V_0 - V_0\right) = \frac{1}{2}CV_0$ $q_2' = C\left(\frac{3}{2}V_0\right) = \frac{3}{2}CV_0$
   • After reaching this state, $S_2$ is released.

3. Step 3: Pressing Switch $S_3$

   • The problem states that $S_2$ is released and then $S_3$ is pressed. The diagram for $S_3$ is ambiguous. A common interpretation in such problems is that $S_3$ connects the two capacitors, now isolated from the batteries, in parallel.
   • Let's assume $S_3$ connects the capacitors in a standard parallel configuration: upper plate to upper plate, and lower plate to lower plate.
   • The two upper plates and their connecting wire are again an isolated system. The total charge will be conserved and will redistribute between the two capacitors.
   • The total charge on the upper plates before pressing $S_3$ is $Q_{total} = q_1' + q_2' = \frac{1}{2}CV_0 + \frac{3}{2}CV_0 = 2CV_0$.
   • When connected in parallel, the charge redistributes until the potential difference across each capacitor is the same. Let the final charges on the upper plates be $q_{1,f}$ and $q_{2,f}$.
   • $V_{f} = \frac{q_{1,f}}{C} = \frac{q_{2,f}}{C} \implies q_{1,f} = q_{2,f}$.
   • The total charge is conserved: $q_{1,f} + q_{2,f} = Q_{total} = 2CV_0$.
   • Substituting $q_{1,f} = q_{2,f}$ into the conservation equation: $q_{1,f} + q_{1,f} = 2CV_0$ $2q_{1,f} = 2CV_0 \implies q_{1,f} = CV_0$
   • Since $q_{1,f} = q_{2,f}$, we also have $q_{2,f} = CV_0$.
   • So, the final charge on the upper plate of $C_1$ is $CV_0$, and the final charge on the upper plate of $C_2$ is also $CV_0$.

Evaluation of Options:

• A: The charge on the upper plate of ${C_1}$ is $2C{V_0}$. This is false. The final charge is $CV_0$.
• B: The charge on the upper plate of ${C_1}$ is $C{V_0}$. This is true based on our calculation.
• C: The charge on the upper plate of ${C_2}$ is $0$. This is false. The final charge is $CV_0$.
• D: The charge on the upper plate of ${C_2}$ is $- C{V_0}$. This is false. The final charge is $CV_0$.

Conclusion:

Based on the most plausible interpretation of the circuit diagram and the sequence of events, only option B is correct. The stored answer indicates that D is also correct. This suggests a potential error in the problem statement, the diagram's representation of the connection made by $S_3$, or the provided correct answer. For instance, if the connection made by $S_3$ were to connect the capacitors in parallel but with opposite polarity, the final charges would be $q_{1,f} = -0.5CV_0$ and $q_{2,f} = 0.5CV_0$, which does not match any option. Given the options, the parallel connection with the same polarity is the most likely intended scenario, which makes B correct. The incorrectness of D under this interpretation points to an error in the question or its provided answer key.