Two identical parallel plate capacitors are connected to a battery...

Initially, when the switch $S$ is closed, both capacitors $A$ and $B$ are connected in parallel to the battery $V$.
Initial energy stored in capacitor $A$, $U_{Ai} = \dfrac{1}{2}CV^2$
Initial energy stored in capacitor $B$, $U_{Bi} = \dfrac{1}{2}CV^2$
Total initial energy, $U_i = U_{Ai} + U_{Bi} = CV^2$

When the switch $S$ is opened, capacitor $A$ remains connected to the battery, so its potential difference remains $V$. Capacitor $B$ is disconnected from the battery, so its charge remains constant at $Q = CV$.
A dielectric of constant $K = 3$ is inserted into both capacitors, so their new capacitances become $C_A' = 3C$ and $C_B' = 3C$.

Final energy stored in capacitor $A$ (voltage is constant):
$U_{Af} = \dfrac{1}{2}C_A'V^2 = \dfrac{1}{2}(3C)V^2 = \dfrac{3}{2}CV^2$

Final energy stored in capacitor $B$ (charge is constant):
$U_{Bf} = \dfrac{Q^2}{2C_B'} = \dfrac{(CV)^2}{2(3C)} = \dfrac{1}{6}CV^2$

Total final energy:
$U_f = U_{Af} + U_{Bf} = \dfrac{3}{2}CV^2 + \dfrac{1}{6}CV^2 = \dfrac{10}{6}CV^2 = \dfrac{5}{3}CV^2$

The ratio of the initial total energy to the final total energy is:
$\dfrac{U_i}{U_f} = \dfrac{CV^2}{\dfrac{5}{3}CV^2} = \dfrac{3}{5}$

Thus, the required ratio is $3 : 5$.