Potential energy of two charged particles separated by distance $r$ is given by,

$U=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{r}$.

Applying conservation of mechanical energy,

$$\begin{gathered} ∆U+∆K=0 \\ \Rightarrow U_i-U_f=\frac{1}{2}m{v}^2 \\ \Rightarrow -\frac{qe}{4\pi {\epsilon }_0}\left(\frac{1}{r_f}-\frac{1}{r_i}\right)=\frac{1}{2}m{v}^2 \\ \Rightarrow v=\sqrt{\frac{2}{9\times {10}^{-31}}\times 20\times {10}^{-9}\times 1.6\times {10}^{-19}\times 9\times {10}^9\times \frac{1}{4}}=4\times {10}^6 \mathrm{m} {\mathrm{s}}^{-1} \end{gathered}$$