The charge on the capacitor in steady state in the circuit shown in the figure is

$I_1=\frac{5}{2+3}=1\mathrm{A}$
$\begin{array}{l}{\mathrm{I}}_2=\frac{5}{3+3}=\frac{5}{6}\mathrm{A}\\ {\mathrm{V}}_{\mathrm{A}}-0=3{\mathrm{I}}_1\text{ or }{\mathrm{V}}_{\mathrm{A}}=3\mathrm{V}\\ {\mathrm{V}}_{\mathrm{B}}-0=3{\mathrm{I}}_2\text{ or }{\mathrm{V}}_{\mathrm{B}}=2.5\mathrm{V}\end{array}$
Potential difference across capacitor is
$\begin{array}{r}3\mathrm{V}-2.5\mathrm{V}=0.5\mathrm{V}=\frac{1}{2}\mathrm{V}\\ \mathrm{Q}=\mathrm{CV}=2\times 1/2=1\mathrm{\mu C}\end{array}$