Answer
$F = 1.94 \times 10^{-5} \mathrm{N} $
Work Step by Step
The kinetic energy for the two protons equals the potential energy at the maximum value
\begin{gather*}
2 K=U \\
2\left(\frac{1}{2} m V^{2}\right)= \frac{1}{4 \pi \epsilon_{0}} \frac{\left|qq\right|}{r}\\
m V^2 = \frac{1}{4 \pi \epsilon_{0}} \frac{\left|e^{2}\right|}{r}\\
r =\frac{1}{4 \pi \epsilon_{0}} \frac{e^{2}}{m V^{2}}\\
r =\left(9.0 \times 10^{9} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}\right) \frac{\left(1.602 \times 10^{-19} \mathrm{C}\right)^{2}}{\left(1.67 \times 10^{-27} \mathrm{kg}\right)\left(2.00 \times 10^{5} \mathrm{m} / \mathrm{s}\right)^{2}}\\
r= 3.45 \times 10^{-12} \mathrm{m}
\end{gather*}
The maximum electric force between the two protons is obtained at $r$ by
\begin{aligned}
F_{max} &=\frac{1}{4 \pi \epsilon_{\mathrm{o}}} \frac{\left|e^{2}\right|}{r^{2}} \\
&=\left(9.0 \times 10^{9} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}\right) \frac{\left|\left(1.602 \times 10^{-19} \mathrm{C}\right)^{2}\right|}{\left(3.45 \times 10^{-12} \mathrm{m}\right)^{2}} \\
&=\boxed{1.94 \times 10^{-5} \mathrm{N} }
\end{aligned}