Answer
$-5.1\times10^{-19}\;\rm J$
Work Step by Step
The potential energy of this system is given by
$$U_{\rm sys}=U_{12}+U_{23}+U_{13}+U_{14}+U_{24}+U_{34}\tag 1$$
We can see, from the figure below, that
$$U_{12}=U_{23}=U_{13}=\dfrac{k(-e)(-e)}{r}=\dfrac{ke^2}{r}\tag 2$$
and that
$$U_{14}=U_{24}=U_{34}=\dfrac{k(-e)(+e)}{r'}=\dfrac{-ke^2}{r'}$$
where $r'=\dfrac{0.5r}{\cos30^\circ}=\dfrac{r}{\sqrt{3}}$
$$U_{14}=U_{24}=U_{34}=\dfrac{k(-e)(+e)}{r'}=\dfrac{-\sqrt{3}ke^2 }{r}\tag 3$$
From (1);
$$U_{\rm sys}=3 U_{12} +3U_{14} $$
Plug from (2), and (3),
$$U_{\rm sys}=3 \left[\dfrac{ke^2}{r}\right]+3\left[ \dfrac{-\sqrt{3}ke^2 }{r} \right] $$
$$U_{\rm sys}= \dfrac{3ke^2}{r} + \dfrac{-3\sqrt{3}ke^2 }{r} $$
$$U_{\rm sys}= \dfrac{3ke^2}{r}\left[1 - \sqrt{3} \right] $$
$$U_{\rm sys}= \dfrac{3(9\times 10^9)(1.6\times 10^{-19})^2}{1\times 10^{-9}}\left[1 - \sqrt{3} \right] $$
$$U_{\rm sys}=\color{red}{\bf -5.1\times10^{-19}}\;\rm J$$
