Answer
$W=2.30\times 10^{-30}J$
Work Step by Step
The initial electric energy is given as
$U_i=\frac{Kq^2}{a}$
The final electric energy is given as
$U_f=3\times \frac{Kq^2}{a}$
We know that
$W=\Delta U=U_f-U_i$
We plug in the known values to obtain:
$W=3\times \frac{Kq^2}{a}-\frac{Kq^2}{a}=2\times \frac{Kq^2}{a}$
We plug in the known values to obtain:
$W=2\times \frac{9\times 10^9(1.6\times 10^{-19})^2}{2\times 10^{-6}}=2.30\times 10^{-30}J$