A source of potential difference V is connected to the combination...

(i) When the switch is closed ${C}_{eq} = 2C$

Charge on each capacitor will be $q = CV$.

Energy ${E}_{1} = \dfrac{1}{2}{C}_{eq}{V}^{2}$ $= \dfrac{1}{2}2C \times {V}^{2}$

${E}_{1} = C{V}^{2}$

(ii) When the switch is opened charge on the right capacitor remain $CV$ while the potential on the left capacitor remains the same.

Dielectric $k = 5$

${C}^{'} = kC$ ${C}^{'} = 5C$

Now to calculate ${E}_{2}$,

${E}_{2} = \dfrac{1}{2}{C}^{'}{V}^{2} + \dfrac{{q}^{2}}{2{C}^{'}}$

$⇒ {E}_{2} = \dfrac{1}{2}\left(5C\right){V}^{2} + \dfrac{{\left(CV\right)}^{2}}{2\left(5C\right)}$

$⇒ {E}_{2} = \dfrac{5C{V}^{2}}{2} + \dfrac{C{V}^{2}}{10}$

$⇒ {E}_{2} = \dfrac{13C{V}^{2}}{5}$

$⇒ \dfrac{{E}_{1}}{{E}_{2}} = \dfrac{5}{13}$