Explanation

(c) Combined capacity of $1\ \mu F$ and $5\mu F\ $= 1 + 5=$6\ \mu F$
Now, $4\mu F\ $and $6\mu F$ are in series.
$\therefore\ \ \frac{1}{C_{s}} = \frac{1}{4} + \frac{1}{6} + \frac{3 + 2}{12} = \frac{5}{12}$
$C_{s} = \frac{12}{5}\mu F$
Charge in the arm containing $\ 4\mu F\ $capacitor is
$q = C_{s} \times V = \frac{12}{5} \times 10 = 24\ \mu C$