Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that the electric potential of a charged sphere on its surface is given by
$$V=\dfrac{1}{4\pi \epsilon_0}\dfrac{q}{R}\tag 1$$
Recalling that $U=qV$, so
$$dU=dq V$$
Plug $V$ from (1),
$$\boxed{dU=\dfrac{1}{4\pi \epsilon_0}\dfrac{q}{R}dq}$$
This is the infinitesimal increase in electric potential energy.
$$\color{blue}{\bf [b]}$$
We just need to integrate the boxed formula above.
$$\int_0^UdU=\int_0^Q \dfrac{1}{4\pi \epsilon_0}\dfrac{q}{R}dq$$
$$U=\dfrac{1}{(4\pi \epsilon_0)R}\int_0^Qq dq=\dfrac{1}{2(4\pi \epsilon_0)R} q^2\bigg|_0^Q$$
$$\boxed{U= \dfrac{1}{(4\pi \epsilon_0) } \dfrac{Q^2}{2R}}$$
$$\color{blue}{\bf [c]}$$
We just need to plug the given into the previous boxed formula,
$$U=(9\times 10^9)\dfrac{(1.6\times 10^{-19})^2}{(1\times 10^{-15)}}$$
$$U=\color{red}{\bf 2.3\times 10^{-13}}\;\rm J$$
