Explanation

(c) Electrostatic potential energy $= \frac{1}{4\pi\varepsilon_{0}}\frac{q_{1}q_{2}}{d}$
$\therefore Energy = \frac{(9 \times 10^{9}) \times ( + 1.6 \times 10^{- 19}) \times ( - 1.6 \times 10^{- 19})J}{10^{- 10}}$
$= - 9 \times 10^{9} \times 1.6 \times 10^{- 19} \times 10^{10}eV = \ - 14.4eV$
Note : In the solution given all the values are positive. It is important to mention the sign