The amount of work done in increasing voltage across the plates of capacitor from 5 V to 10 V is W. The work done in increasing it from 10 V to 20 V is

The work done in increasing the voltage across a capacitor is equal to the change in its electrostatic potential energy.

$W = \Delta U = \dfrac{1}{2} C (V_f^2 - V_i^2)$

For the first case, the voltage increases from $5$ V to $10$ V:

$W = \dfrac{1}{2} C (10^2 - 5^2) = \dfrac{1}{2} C (100 - 25) = \dfrac{75}{2} C$

For the second case, the voltage increases from $10$ V to $20$ V:

$W' = \dfrac{1}{2} C (20^2 - 10^2) = \dfrac{1}{2} C (400 - 100) = \dfrac{300}{2} C$

Taking the ratio of the two work done values:

$\dfrac{W'}{W} = \dfrac{300}{75} = 4$

$W' = 4W$

Answer: $4W$