Answer
$U = 2.42\times 10^{-29}~J$
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 distance $d$ between $q_1$ and $q_2$:
$d = \sqrt{d_1^2+d_2^2}$
$d = \sqrt{(4.00~m)^2+(2.00~m)^2}$
$d = 4.47~m$
We can find the electric potential energy of the three-particle system:
$U = \frac{1}{4\pi~\epsilon_0}~(\frac{q_1~q_3}{d_1}+ \frac{q_2~q_3}{d_2}+ \frac{q_1~q_2}{4.47~m})$
$U = \frac{1}{(4\pi)~(8.854\times 10^{-12}~C/V~m)}~[\frac{(-2)(1.6\times 10^{-19}~C)(2)(1.6\times 10^{-19}~C)}{4.00~m}+ \frac{(2)(1.6\times 10^{-19}~C)(2)(1.6\times 10^{-19}~C)}{2.00~m}+ \frac{(-2)(1.6\times 10^{-19}~C)(2)(1.6\times 10^{-19}~C)}{4.47~m}]$
$U = 2.42\times 10^{-29}~J$
