Answer
$U = 97 \times 10^{-3} \,\text{J}$
Work Step by Step
There are three charges and each charge $q$ is at distance $r$ = 0.4 m from the other two. So, the energy that assembles the three charges will be the summation of the potential energy of each charge and will be calculated in the form
\begin{align*}
U &= \dfrac{1}{4\pi \epsilon_o} \sum \dfrac{q q_i}{r_i}\\
&= \dfrac{1}{4\pi \epsilon_o} \dfrac{+3q^2}{r}\\
&=(9\times 10^{9} \mathrm{~N\cdot m^2/C^2}) \dfrac{+3(1.2\times 10^{-6} \,\text{C})^2}{0.4 \,\text{m}}\\
&= \boxed{97 \times 10^{-3} \,\text{J}}
\end{align*}
