Explanation

(d) In steady state current through the branch having capacitor is zero.
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{R} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{R} = \frac{6 + 3 + 2}{6}$
$R = \frac{6}{11}$
As $V = iR$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 6 = i \times \frac{6}{11}$
Current through the battery $i = 11A$
Charge on the capacitor $q = CV$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ q = 0.5 \times 10^{- 6} \times 6$
$q = 3\mu C$