The provided solution requires interpreting the circuit diagram in a specific way that might not be immediately obvious. Based on the options, the intended configuration is likely that the capacitors are in series. Let's solve the problem with this assumption.

Assumed Circuit Configuration: The 3 µF capacitor (C1) and the 6 µF capacitor (C2) are connected in series across the 9V battery. Point Y is the junction between the two capacitors. Point X is connected to the negative terminal of the battery. The switch S connects points Y and X.

Let's set the potential of the negative terminal of the battery (and point X) to be 0 V. Then the potential of the positive terminal is 9 V.

Step 1: Initial State (Switch S is open)

1. Equivalent Capacitance: When two capacitors are in series, the equivalent capacitance ($C_{eq}$) is given by: $C_{eq} = \frac{C_1 C_2}{C_1 + C_2} = \frac{(3 \mu F)(6 \mu F)}{3 \mu F + 6 \mu F} = \frac{18}{9} \mu F = 2 \mu F$

2. Total Charge: The total charge drawn from the battery and stored on the series combination is: $Q = C_{eq} V = (2 \mu F)(9 V) = 18 \mu C$ In a series combination, this is the magnitude of the charge on each capacitor. $Q_1 = Q_2 = 18 \mu C$.

3. Potential Distribution: The potential difference across each capacitor is: $V_1 = \frac{Q}{C_1} = \frac{18 \mu C}{3 \mu F} = 6 V$ $V_2 = \frac{Q}{C_2} = \frac{18 \mu C}{6 \mu F} = 3 V$ (Check: $V_1 + V_2 = 6 V + 3 V = 9 V$, which is the battery voltage).

4. Initial Potentials and Charges:

   • Potential at X: $V_X = 0 V$.
   • Potential at Y: Point Y is between C1 and C2. Its potential is the voltage across C2 relative to point X. $V_Y = V_X + V_2 = 0 V + 3 V = 3 V$.
   • Charge on the plates connected to Y:
     • The plate of C1 connected to Y is at a lower potential (3V) than its other plate (9V), so its charge is negative: $q_{1Y, initial} = -18 \mu C$.
     • The plate of C2 connected to Y is at a higher potential (3V) than its other plate (0V), so its charge is positive: $q_{2Y, initial} = +18 \mu C$.

Step 2: Final State (Switch S is closed)

1. Final Potentials: When the switch S is closed, it connects points Y and X. This forces them to be at the same potential. Since $V_X = 0 V$, the final potential of Y is $V_{Y, final} = 0 V$.

2. Final Charges:

   • Capacitor C1 (3 µF): It is now connected between the positive terminal (9 V) and point Y (0 V). The potential difference across it is $9 V - 0 V = 9 V$. The final charge on C1 is $Q'_1 = C_1 (9 V) = (3 \mu F)(9 V) = 27 \mu C$. The plate connected to Y is at the lower potential (0V), so its final charge is $q_{1Y, final} = -27 \mu C$.
   • Capacitor C2 (6 µF): It is now connected between point Y (0 V) and point X (0 V). The potential difference across it is 0 V. The final charge on C2 is $Q'_2 = 0$. The final charge on the plate connected to Y is $q_{2Y, final} = 0$.

Step 3: Calculate Charge Flow

We need to find the total charge that flows from Y to X through the switch. Let this be $Q_{Y \to X}$. We can find this by applying the principle of charge conservation at node Y.

The sum of charges flowing out from node Y must be zero. The charge flows out from Y into the plates of the capacitors and through the switch.

Charge flow from Y to C1's plate = Change in charge of C1's plate connected to Y. $Q_{Y \to C1} = \Delta q_{1Y} = q_{1Y, final} - q_{1Y, initial} = (-27 \mu C) - (-18 \mu C) = -9 \mu C$

Charge flow from Y to C2's plate = Change in charge of C2's plate connected to Y. $Q_{Y \to C2} = \Delta q_{2Y} = q_{2Y, final} - q_{2Y, initial} = (0 \mu C) - (18 \mu C) = -18 \mu C$

Applying charge conservation at node Y: $Q_{Y \to C1} + Q_{Y \to C2} + Q_{Y \to X} = 0$ $(-9 \mu C) + (-18 \mu C) + Q_{Y \to X} = 0$ $-27 \mu C + Q_{Y \to X} = 0$ $Q_{Y \to X} = 27 \mu C$

Thus, the total amount of charge that flows from Y to X is 27 µC.