Easy problem from Electrostatics

Let A and B be two points at distance 10cm and 5cm from the charge.
$V_{A} and V_{B}$ are values of potential at these points.
Now, $V_{A}$ = $\frac{1}{4\pi\epsilon}*\frac{q}{r_{1}}=9*10^{9}*\frac{20*10^{-6}}{0.1}=1.8*10^{6}V$
Similarly if we calculate $V_{B}$ it comes out to be $3.6*10^{6}V$
Difference between $V_{A} and V_{B}$ is $1.8*10^6$ V.
Work done to take the electron from A to B is $1.8*10^{6}*1.6*10^{-19}=\boxed{2.88*10^{-13}J}$