1. Initial charge on $C_1$

When $C_1 = 1\,\mu F$ is connected to the battery of voltage $60\,V$, it gets charged to

$Q_0 = C_1 V = (1\,\mu F)(60\,V) = 60\,\mu C$

So initially, capacitor $C_1$ has charge $60\,\mu C$.

2. Connection after switching

After disconnecting from the battery, $C_1$ is connected to the combination of uncharged capacitors $C_2 = 3\,\mu F$ and $C_3 = 6\,\mu F$.

From the figure/context, the three capacitors are connected in parallel after switch (2), so they attain a common final voltage $V_f$.

Since the battery is removed, total charge is conserved.

Thus,

$Q_{\text{total}} = 60\,\mu C$

Total capacitance in parallel:

$C_{\text{eq}} = C_1 + C_2 + C_3 = 1 + 3 + 6 = 10\,\mu F$

Hence the common final voltage is

$V_f = \frac{Q_{\text{total}}}{C_{\text{eq}}} = \frac{60\,\mu C}{10\,\mu F} = 6\,V$

3. Final charges on $C_2$ and $C_3$

For $C_2$:

$Q_2 = C_2 V_f = (3\,\mu F)(6\,V) = 18\,\mu C$

For $C_3$:

$Q_3 = C_3 V_f = (6\,\mu F)(6\,V) = 36\,\mu C$

Sum of final charges on $C_2$ and $C_3$:

$Q_2 + Q_3 = 18 + 36 = 54\,\mu C$

4. Check options

• A: $40\,\mu C$ ❌
• B: $36\,\mu C$ ❌
• C: $20\,\mu C$ ❌
• D: $54\,\mu C$ ✅

Therefore, the correct answer is:

$\boxed{54\,\mu C}$