The system of two protons and one electron is represented in the given figure.

Charge on proton $1, q_{1}=1.6 \times 10^{-19} \,C$

Charge on proton $2, q_{2}=1.6 \times 10^{-19} \,C$

Charge on electron, $q_{3}=-1.6 \times 10^{-19} \,C$

Distance between protons $1$ and $2, d _{1}=1.5 \times 10^{-10} \,m$

Distance between proton $1$ and electron, $d _{2}=1 \times 10^{-10}\, m$

Distance between proton $2$ and electron, $d _{3}=1 \times 10^{-10} \,m$

The potential energy at infinity is zero. Potential energy of the system,

$V=\frac{q_{1} q_{2}}{4 \pi \epsilon_{0} d_{1}}+\frac{q_{2} q_{3}}{4 \pi \epsilon_{0} d_{3}}+\frac{q_{1} q_{1}}{4 \pi \epsilon_{0} d_{2}}$

Substituting $\frac{1}{4 \pi \epsilon_{0}}=9 \times 10^{9} \,N\,m ^{2} \,C ^{-2},$ we obtain

$V =\frac{9 \times 10^{9} \times 10^{-19} \times 10^{-19}}{10^{-10}}\left[-(16)^{2}+\frac{(1.6)^{2}}{1.5}+-(1.6)^{2}\right]$

$=-30.7 \times 10^{-19} \,J$

$=-19.2\, eV$

Therefore, the potential energy of the system is $-19.2\; eV.$