Answer
The amount of work done is $~~2.4~J$
Work Step by Step
We can find the initial amount of energy stored in the capacitor:
$E_1 = \frac{Q^2}{2C_1}$
$E_1 = \frac{(4.0\times 10^{-3}~C)^2}{(2)(5.0\times 10^{-6}~F)}$
$E_1 = 1.6~J$
We can find the final amount of energy stored in the capacitor:
$E_2 = \frac{Q^2}{2C_2}$
$E_2 = \frac{(4.0\times 10^{-3}~C)^2}{(2)(2.0\times 10^{-6}~F)}$
$E_2 = 4.0~J$
The amount of work done by the external force is equal to the change in energy.
The amount of work done is $4.0~J - 1.6~J$ which is $2.4~J$
