Three identical capacitors C_1 , C_2 and C_3 , each of capacitance...

The equivalent capacitance of the parallel combination of $C_1$ and $C_2$ is:
$C_{12} = C + C = 2C$

This combination is connected in series with $C_3$ (capacitance $C$) across the voltage source $V$. In a series circuit, the voltage divides inversely as the capacitance. The voltage across the parallel combination (which is the voltage across $C_1$) is:
$V_1 = V \times \dfrac{C_3}{C_{12} + C_3} = V \times \dfrac{C}{2C + C} = \dfrac{V}{3}$

The voltage across $C_3$ is:
$V_3 = V \times \dfrac{C_{12}}{C_{12} + C_3} = V \times \dfrac{2C}{2C + C} = \dfrac{2V}{3}$

The energy stored in capacitor $C_1$ is:
$U_1 = \dfrac{1}{2} C_1 V_1^2 = \dfrac{1}{2} C \left(\dfrac{V}{3}\right)^2 = \dfrac{CV^2}{18}$

The energy stored in capacitor $C_3$ is:
$U_3 = \dfrac{1}{2} C_3 V_3^2 = \dfrac{1}{2} C \left(\dfrac{2V}{3}\right)^2 = \dfrac{4CV^2}{18}$

The ratio of the energy stored in $C_1$ to that in $C_3$ is:
$\dfrac{U_1}{U_3} = \dfrac{\dfrac{CV^2}{18}}{\dfrac{4CV^2}{18}} = \dfrac{1}{4}$

Answer: $1:4$