Answer
$ U = 1.15 GeV $
Work Step by Step
For $^{230}Pu $ nuclide where $r= 6.64 fm$ and $q = 94e$
$U = \frac{3q^2}{20 \pi \epsilon_0 r }$
$U = \frac{3(94e)^2}{20 \pi (8.85 \times 10^{-12} F/m) (6.64 \times 10^{-15} m) }$
$U = 1.8425 \times 10^{-10} J$
Change unit to eV,
$ U = 1.8425 \times 10^{-10} J \times \frac{1 eV}{ 1.602 \times 10^{-19} J}$
$ U = 1.15 GeV $
