d
(d)
When capacitor is fully charged, it acts like an open switch or path. So, we have following equivalent circuit
Clearly, $\quad V_{C D}=V$
So, $V_{A B}=I_{A B} \cdot R_{A B}$
$=\frac{V}{\left(R+\frac{5}{3} R\right)} \cdot\left(\frac{5}{3} R\right)=\frac{5}{8} \,V$
Charge stored on the capacitor is
$Q=C_{A B} \cdot V_{A B}=\frac{5}{8} \,C V$
