Applying Kirchhoff's law

For loop $ABEFGA:$ $R i_{1}+R\left(i_{1}+i_{2}\right)=15$

$\Longrightarrow 2 i_{1}+i_{2}=\frac{15}{R}=\frac{15}{3}=5 \ldots .(1)$

For loop $ABCDEFGA:$ $R i_{2}+R i_{2}+R\left(i_{1}+i_{2}\right)=15$

$\Longrightarrow i_{1}+3 i_{2}=\frac{15}{R}=\frac{15}{3}=5 \ldots . .(2)$

From $(1)$ and $( 2 ):$

$i_{1}+3\left(5-2 i_{1}\right)=5$

$\Rightarrow 5 i_{1}=10$

$\Rightarrow i_{1}=2 A$

Putting, $i_{1}=2$ in $(1)$

$\Rightarrow i_{2}=5-2(2)=1 A$

Thus, Potential across capacitor is: $V_{c}=R i_{2}+R\left(i_{1}+i_{2}\right)=(3 \times 1)+3(2+1)=3+$

$9=12 V$