$C _{ eq }=2 C$

Energy $E_{1}=\frac{1}{2} C_{\text {oq }} V^{2}$

$=\frac{1}{2} 2 C ^{2} \times V ^{2}$

$E _{1}= CV ^{2}$

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

Dielectric $K =5$

$C ^{\prime}= KC$

$C ^{\prime}=5\,C$

$E _{2}=\frac{1}{2}(5 C ) V ^{2}+\frac{( CV )^{2}}{2(5 C )}$

$E _{2}=\frac{5 CV ^{2}}{2}+\frac{ CV ^{2}}{10}$

$E _{2}=\frac{13 CV ^{2}}{5}$

$\frac{ E _{1}}{ E _{2}}=\frac{ CV ^{2}}{\frac{13 CV ^{2}}{5}}=\frac{5}{13}$

$\frac{ E _{1}}{ E _{2}}=\frac{5}{13}$