Answer
The electric potential energy decreases by 36%
Work Step by Step
For the solution,
Consider,
$U_{f}$ = Final Potential Energy which consists of arrangement of Xenon and Strontium
$U_{i}$ = Initial Potential Energy which consists of Uranium
$R_{Xe}$, $R_{Sr}$ & $R_{U}$ to be radius of Xenon, Strontium and Uranium atoms whereas
$Q_{Xe}$, $Q_{Sr}$ & $
Q_{U}$ to be charge of Xenon, Strontium and Uranium atoms
The fractional change in potential energy is
$$
\begin{aligned}
\frac{\Delta U}{U} &=\frac{U_{f}}{U_{i}}-1=\frac{Q_{\mathrm{Xe}}^{2} / R_{\mathrm{Xe}}+Q_{\mathrm{Sr}}^{2} / R_{\mathrm{Sr}}}{Q_{\mathrm{U}}^{2} / R_{\mathrm{U}}}-1=\frac{(54)^{2}(140)^{-1 / 3}+(38)^{2}(96)^{-1 / 3}}{(92)^{2}(236)^{-1 / 3}}-1 \\
&=-0.36
\end{aligned}
$$
The percentage change in potential energy is given by ;
$\dfrac{\Delta U}{U}$ x 100 = -0.36 X 100 = -36%
