When battery is replaced by another uncharged capacitor

As uncharged capacitor is connected parallel

So, $\quad \mathrm{C}^{\prime}=2 \mathrm{C}$

and $\quad \mathrm{V}_{\mathrm{c}}=\frac{\mathrm{q}_{1}+\mathrm{q}_{2}}{\mathrm{C}_{1}+\mathrm{C}_{2}}$

$\mathrm{V}_{\mathrm{c}}=\frac{\mathrm{q}+0}{\mathrm{C}+\mathrm{C}} \Rightarrow \mathrm{V}_{\mathrm{c}}=\frac{\mathrm{V}}{2}$

Initial Energy of system, $\mathrm{U}_{\mathrm{i}}=\frac{1}{2} \mathrm{CV}^{2} \ldots(\mathrm{i})$

Final energy of system, $\mathrm{U}_{\mathrm{f}}=\frac{1}{2}(2 \mathrm{C})\left(\frac{\mathrm{V}}{2}\right)^{2}$

$=\frac{1}{2} \mathrm{CV}^{2}\left(\frac{1}{2}\right)…..(ii)$

From equation $(i)$ and $(ii)$

$\mathrm{U}_{\mathrm{f}}=1 / 2 \mathrm{U}_{\mathrm{i}}$

i.e., Total electrostatic energy of resulting system decreases by a factor of $2$