Explanation

(b) Energy of a charged capacitor,
$\ \ \ \ E = \frac{1}{2}\frac{Q^{2}}{C}$
$\ \ \ C = \ \frac{2\pi\varepsilon_{0}L}{\log_{e}\left( \frac{b}{a} \right)}$
$\ E^{'} = \ \frac{1}{2}\frac{Q^{2}}{2\pi\varepsilon_{0}L}\log_{e}\left( \frac{b}{a} \right)\ $ …(i)
For a cylindrical capacitor.
where $L$= length of the cylinders
$a\ and\ b$ = radii of two concentric cylinders
$\ \ \ \ \ \ \ \ \ \ \ C^{'} = \ \frac{2\pi\varepsilon_{0}(2L)}{\log_{e}\left( \frac{b}{a} \right)}$
$\ \ \ \ \ \ \ \ \ \ E^{'} = \frac{1}{2}\frac{(2Q)^{2}}{C'}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{1}{2}\frac{(2Q)^{2}}{2\pi\varepsilon_{0}(2L)}\log_{e}\left( \frac{b}{a} \right)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $ ...(ii)
From Eqs. (i) and (ii), we get
$\ \ \ \ \ \ \ \ \ \ E^{'} = \ 2E$