Answer
$U = 4.8\times 10^{-6}~J$
Work Step by Step
We can find the electric potential energy:
$U = \frac{kq_1q_2}{r_{12}}+\frac{kq_1q_3}{r_{13}}+\frac{kq_2q_3}{r_{23}}$
$U = k~(\frac{q_1q_2}{r_{12}}+\frac{q_1q_3}{r_{13}}+\frac{q_2q_3}{r_{23}})$
$U = (9.0\times 10^9~N~m^2/C^2)~[\frac{(2.0\times 10^{-9}~C)(3.0\times 10^{-9}~C)}{0.030~m}+\frac{(2.0\times 10^{-9}~C)(3.0\times 10^{-9}~C)}{0.040~m}+\frac{(3.0\times 10^{-9}~C)(3.0\times 10^{-9}~C)}{0.050~m}~]$
$U = 4.8\times 10^{-6}~J$
