Before inserting dielectric capacitance is given C0=12.5pF and charge on the capacitor Q=C0 V After inserting dielectric capacitance will become ϵrC0. Change in potential energy of the capacitor $\begin{aligned}
& =\mathrm{E}{\mathrm{i}}-\mathrm{E}{\mathrm{f}} \
& =\frac{\mathrm{Q}^2}{2 \mathrm{C}{\mathrm{i}}}-\frac{\mathrm{Q}^2}{2 \mathrm{C}{\mathrm{f}}}=\frac{\mathrm{Q}^2}{2 \mathrm{C}0}\left[1-\frac{1}{\epsilon{\mathrm{r}}}\right] \
& =\frac{\left(\mathrm{C}_0 \mathrm{V}\right)^2}{2 \mathrm{C}0}\left[1-\frac{1}{\epsilon{\mathrm{r}}}\right]=\frac{1}{2} \mathrm{C}_0 \mathrm{V}^2\left[1-\frac{1}{\epsilon_{\mathrm{r}}}\right]
\end{aligned}$
Using C0=12.5pF,V=12 V,ϵr=6 $\begin{aligned}
& =\frac{1}{2}(12.5) \times 12^2\left[1-\frac{1}{6}\right]=\frac{1}{2}(12.5) \times 12^2 \times \frac{5}{6} \
& =750 \mathrm{pJ}=750 \times 10^{-12} \mathrm{~J}
\end{aligned}$