The electric potential energy of a pair of point charges $q_l$ $q_2$ and at a distance $r$ from each other can be expressed as $U=\frac{k{q}_1{q}_2}{r_{12}}$
If the charges are of different signs as in the case at hand, then $U$ is negative and negative work is done by an external agent against the attractive force between the charges of opposite signs as they are brought near each other; a force must be applied opposite the displacement to prevent $q_1$ (the electron) from accelerating toward $q_2$ (the proton).
If the system consists of more than two charged particles, we can obtain the total potential energy of the system by calculation $U$ for every pair of charges and summing the terms algebraically.
i.e. $U=k(\frac{q_1{q}_2}{r_{12}}+\frac{q_1{q}_3}{r_{13}}+\frac{q_2{q}_3}{r_{23}}+\dots )$
$$\begin{array}{rl}& U=\frac{q_1{q}_2}{4\pi {\epsilon }_{0}d}\\ & =\frac{1.6\times {10}^{-19}}{4\times 3.142\times 8.85\times {10}^{-12}}\times \frac{1.6\times {10}^{-19}}{0.5\times {10}^{-10}}\end{array}$$
$$\begin{array}{rl}& =\frac{1.6\times {10}^{-19}\times 1.6\times {10}^{-19}}{5.56134\times {10}^{-2b}}\\ & =\frac{2.56\times {10}^{-38}}{5.56134\times {10}^{-21}}=4.6\times {10}^{-18}J\end{array}$$