In the given circuit with steady current the potential drop across the capacitor must be

Apply Kirchhoff’s law to the closed mesh ADFHA, $2\mathrm{V}-\mathrm{I}\left(2\mathrm{R}\right)-\mathrm{I}\left(\mathrm{R}\right)-\mathrm{V}=0$ 

$\mathrm{V}=3\mathrm{IR}\text{ (or) }\mathrm{I}=\frac{\mathrm{V}}{3\mathrm{R}}$ 

Apply Kirchhoff’s law of the mesh ABGHA,

$\mathrm{V}+{\mathrm{V}}_{\mathrm{C}}-\mathrm{IR}-\mathrm{V}=0\text{ (or) }{\mathrm{V}}_{\mathrm{C}}=\mathrm{IR}\therefore {\mathrm{V}}_{\mathrm{C}}=\left(\frac{\mathrm{V}}{3\mathrm{R}}\right)\mathrm{R}=\frac{\mathrm{V}}{3}$

$\therefore$ Potential difference across capacitor is ${\mathrm{V}}_{\mathrm{C}}=\frac{\mathrm{V}}{3}$