$\left(1.6 \times 10^{-19}\right)\left(10^6\right)=\frac{1}{2}\left(1.67 \times 10^{-27}\right) v^2$
$1.6 \times 10^{-13}=\frac{1.67}{2} \times 10^{-27} v^2= KE$
$KE =1.6 \times 10^{-13} \,J$






Note: $V_{1,2,3,4}$ are the potential differences across $C_{1,2,3,4}$ and $Q_{1,2,3,4}$ are the final charges stored in $C_{1,2,3,4}$ respectively.