In the given circuit [fig]

potential difference across the capacitor must be

In steady state no current will pass through the capacitor. The current flowing in the circuit is shown in fig

From fig
$\begin{array}{l}\mathrm{i}=\frac{2\mathrm{V}-\mathrm{V}}{\mathrm{R}+2\mathrm{R}}=\frac{\mathrm{V}}{3\mathrm{R}}\\ {\mathrm{V}}_{\mathrm{C}}-{\mathrm{V}}_{\mathrm{A}}=\mathrm{iR}+\mathrm{V}=\frac{\mathrm{V}}{3}+\mathrm{V}=\frac{4\mathrm{V}}{3}\end{array}$
Further, ${\mathrm{V}}_{\mathrm{C}}-{\mathrm{V}}_{\mathrm{A}}=\left({\mathrm{V}}_{\mathrm{C}}-{\mathrm{V}}_{\mathrm{B}}\right)+\left({\mathrm{V}}_{\mathrm{B}}-{\mathrm{V}}_{\mathrm{A}}\right)$
$\begin{array}{l}\therefore {\mathrm{V}}_{\mathrm{C}}-{\mathrm{V}}_{\mathrm{B}}=\left({\mathrm{V}}_{\mathrm{C}}-{\mathrm{V}}_{\mathrm{A}}\right)-\left({\mathrm{V}}_{\mathrm{B}}-{\mathrm{V}}_{\mathrm{A}}\right)\\ =\frac{4\mathrm{V}}{3}-\mathrm{V}=\frac{\mathrm{V}}{3}\end{array}$
So, the potential drop across the capacitor will be V / 3.