As shown in the given figure, switch is closed for long time. when the switch is open, and then the space between the (both A and B capacitors) capacitors is filled with dielectrics of dielectric constant 3. The ratio of total energy capacitors before and after filling the dielectric will be
 

Initially  $q_A=q_B=CV$
When dielectric is filled, then
   $C_1=K_{1}C=3C (\therefore K_1=3)$
When switch is opened, the p.d. across ‘A’ is V let p.d. across ‘B’ be   
 $$\begin{gathered} \therefore q_B=CV=C_1{V}_{1}orCV=3C{V}_1 \\ \Rightarrow V_1=\frac{V}{3} \end{gathered}$$

Total energy  $E_i=\frac{1}{2}C{V}^2+\frac{1}{2}C{V}^2=C{V}^2$
Final energy of  $A=\frac{1}{2}\left(3C\right)V^2=\frac{3}{2}C{V}^2$
Final energy of  $B=\frac{1}{2}\left(3C\right){\left(\frac{V}{3}\right)}^2=\frac{C{V}^2}{6}$
Total energy  $E_f=\frac{3}{2}C{V}^2+\frac{C{V}^2}{6}=\frac{5}{3}C{V}^2$
 $\therefore \frac{E_i}{E_f}=\frac{C{V}^2}{\frac{5}{3}C{V}^2}=\frac{3}{5}$