a b c d

c

$\text{Equivalent Capacitance of circuit containing } 3\mu F , 2\mu F ,5\mu F is $ $ C_{eq} = \frac{3\times 7\mu F}{3+7}= 2.1 \mu F$ $ \text{Total Charge passing through } 3\mu F = C_{eq} V = 12.6\mu F$ $ \text{Charge on }5\mu F = \frac{5 \times 12.6 \mu F}{7} = 9 \mu F$