Explanation

(c) As the capacitors $4\mu F$ and $2\mu F$ connected in parallel, are in series with $6\mu F$ capacitor, their equivalent capacitance is $\frac{(2 + 4) \times 6}{2 + 4 + 6} = 3\mu F$
Charge in the circuit, $Q = 3\mu F \times 12V = 36\mu C$
Since the capacitors $4\mu F$ and $2\mu F$ are connected in parallel, therefore potential difference across them is same
$\Rightarrow \frac{Q_{1}}{Q_{2}} = \frac{C_{1}}{C_{2}} = \frac{4}{2} \Rightarrow Q_{1} = 2Q_{2}$
Also, $Q = Q_{1} + Q_{2}$
$\therefore 36\mu C = 2Q_{2} + Q_{2} \Rightarrow Q_{2} = \frac{36\mu C}{3} = 12\mu C$
$Q_{1} = Q - Q_{2} = 36\mu C - 12\mu C = 24\mu C = 24 \times 10^{- 6}C$