Answer
$-2.23\times10^{-19}\;\rm J$
Work Step by Step
We know that the electric potential energy is given by
$$U_e=qV\tag 1$$
So we have to find the net electric potential at the proton.
$$V=V_1+V_2=\dfrac{kq_1}{r_1}+\dfrac{kq_2}{r_2}$$
The distance between the upper electron and the proton is equal to the distance between the lower electron and the proton. So $r_1=r_2=r$. And $q_1=q_2=-q$
$$V =\dfrac{-kq}{r }+\dfrac{-kq}{r }=\dfrac{-2kq}{r }$$
Plug into (1),
$$U_e=q\dfrac{-2kq}{r }=\dfrac{-2kq^2}{r }$$
Plug the known and note, from the given graph, that $r=\sqrt{x^2+y^2}=\sqrt{(0.5^2+2^2)\times 10^{-9}}$
$$U_e= \dfrac{-2(9\times 10^9)(1.6\times 10^{-19})^2}{\sqrt{(0.5^2+2^2)\times (10^{-9})^2} }$$
$$U_2=\color{red}{\bf -2.23\times10^{-19}}\;\rm J$$
