b
In steady state current in the branch containing the capacitor is zero and hence
$\mathrm{emf}$ $\mathrm{E}$ is shared between
$\mathrm{r} $ and $ \mathrm{R}_{2}=$ in the ratio of their resistance voltage across $\mathrm{R}_{2}$ is $\frac{\mathrm{ER}_{2}}{\mathrm{R}_{2}+\mathrm{r}}=2 \mathrm{\,Volts}$
$=$ Voltage across capacitor.
$\therefore $ $\mathrm{Q}=\mathrm{CV}_{\mathrm{c}}=2 \mu \mathrm{C}$