d
The capacitor branch will not allow charge to pass through it
$\therefore $ Current in the circuit $\mathrm{I}=\frac{12}{6+2}=\frac{3}{2} \mathrm{\,amp}$
Potential difference across $2\, \mu \mathrm{F}$ is same as across $6\, \Omega$ resistance.
So, potential difference across capacitor of
$2\, \mu \mathrm{F}=6 \times \frac{3}{2}=9$ $\mathrm{volt}.$
$\therefore $ Charge on capacitor $=2 \times 9 \mu C=18\, \mu C$