Answer
2 16 MeV.
Work Step by Step
The required potential is the scalar sum of the individual potential energies. Each individual potential energy is $U = \frac{1}{4\pi \epsilon_{0}} \frac{q_1 q_2}{r}$.
Each charge is $e$ and the charges are all equidistant from one another, so
$\Sigma U = \frac{1}{4\pi \epsilon_{0}}(\frac{e^2}{r} + \frac{e^2}{r} +\frac{e^2}{r}) = \frac{3e^2}{4\pi \epsilon_{0}r}$
Plugging in the data, we have:
$\Sigma U = \frac{3(9\times 10^9)(1.6\times 10^{-19})^2}{(2\times 10^{-15})} = 3.46 \times 10^{-13} J = 2.16 \ MeV$.
