In the circuit shown, the energy stored in the capacitor is nμ J ....

The capacitor works as an open circuit in the given configuration.

The potential difference between the two plates of the capacitor will be the difference across the resistors having resistances $3Ω and 4Ω$.

Hence, the potential difference $\left(∆ V\right)$ across the capacitor can be calculated as follows:

$\begin{cases}\DeltaV & = & \left(\dfrac{4}{6} \times 12 - \dfrac{3}{12} \times 12\right)V \\ & = & 5V\end{cases}$

The formula to calculate the energy stored in the capacitor is given by

$U = \dfrac{1}{2}C ( ∆ V{)}^{2} . . . \left(1\right)$

Substitute the values of the known parameters into equation (1) to calculate the required energy stored.

$\begin{cases}U & = & \dfrac{1}{2} \times 6\mu F \times {\left(5V\right)}^{2} \\ & = & 75\mu J\end{cases}$