Answer
1.3 eV is lost.
Work Step by Step
The potential energy of the two equal but opposite charges is $PE=k\frac{(e)(-e)}{r}$.
Calculate the change in PE as final PE minus initial PE.
$$\Delta PE=-ke^2(\frac{1}{r_f}-\frac{1}{r_i})$$
$$\Delta PE=-(8.988\times10^9)(1.60\times10^{-19})^2(\frac{1}{0.100\times10^{-9}}-\frac{1}{0.110\times10^{-9}})\frac{1eV}{1.60\times10^{-19}J}$$
$$\Delta PE=-1.31eV$$
The amount lost is 1.3 eV.
