Answer
$U = 0$
Work Step by Step
We can write a general expression for the electric potential energy of a system of charged particles:
$U = \frac{1}{4\pi~\epsilon_0}~\sum \frac{q_i~q_j}{r_{ij}}$
We can find the electric potential energy of the three-quark system:
$U = \frac{1}{4\pi~\epsilon_0}~(\frac{q_1~q_2}{d}+\frac{q_1~q_3}{d}+\frac{q_2~q_3}{d})$
$U = \frac{1}{4\pi~\epsilon_0}~[\frac{(2e/3)(2e/3)}{d}+\frac{(2e/3)(-e/3)}{d}+\frac{(2e/3)(-e/3)}{d}]$
$U = \frac{1}{4\pi~\epsilon_0}~[\frac{4e^2}{9d}-\frac{2e^2}{9d}-\frac{2e^2}{9d}]$
$U = 0$
