Two identical air capacitors A and B are connected in series across...

Let the initial capacitance of each air capacitor be $C$.

Initially, capacitors $A$ and $B$ are in series. The equivalent capacitance is:
$C_{eq,i} = \dfrac{C \times C}{C + C} = \dfrac{C}{2}$

The total initial energy stored in the system is:
$U_i = \dfrac{1}{2} C_{eq,i} V^2 = \dfrac{1}{2} \left(\dfrac{C}{2}\right) V^2 = \dfrac{CV^2}{4}$

When a dielectric of $K = 2$ is inserted into capacitor $B$, its new capacitance becomes $C_B' = 2C$. Capacitor $A$ remains $C_A = C$.

The new equivalent capacitance of the series combination is:
$C_{eq,f} = \dfrac{C_A \times C_B'}{C_A + C_B'} = \dfrac{C \times 2C}{C + 2C} = \dfrac{2C}{3}$

Since the battery remains connected, the voltage across the combination is still $V$. The total final energy stored in the system is:
$U_f = \dfrac{1}{2} C_{eq,f} V^2 = \dfrac{1}{2} \left(\dfrac{2C}{3}\right) V^2 = \dfrac{CV^2}{3}$

The ratio of the initial total energy to the final total energy is:
$\dfrac{U_i}{U_f} = \dfrac{\dfrac{CV^2}{4}}{\dfrac{CV^2}{3}} = \dfrac{3}{4}$

Answer: $3/4$