A circuit consists of two parallel branches connected across an 18...

For Branch 1, capacitors $C_1$ and $C_2$ are in series. Their equivalent capacitance is:
$C_{s1} = \dfrac{C_1 C_2}{C_1 + C_2} = \dfrac{2 \times 4}{2 + 4} = \dfrac{8}{6} = \dfrac{4}{3}\ \mu\text{F}$.

The charge on Branch 1 is:
$Q_1 = C_{s1} \times V = \dfrac{4}{3}\ \mu\text{F} \times 18\text{ V} = 24\ \mu\text{C}$.
Since $C_1$ and $C_2$ are in series, the charge on $C_1$ is $24\ \mu\text{C}$.
The electrostatic energy stored in $C_1$ is:
$U_1 = \dfrac{Q_1^2}{2C_1} = \dfrac{(24\ \mu\text{C})^2}{2 \times 2\ \mu\text{F}} = \dfrac{576}{4}\ \mu\text{J} = 144\ \mu\text{J}$.

For Branch 2, capacitors $C_3$ and $C_4$ are in series. Their equivalent capacitance is:
$C_{s2} = \dfrac{C_3 C_4}{C_3 + C_4} = \dfrac{3 \times 6}{3 + 6} = \dfrac{18}{9} = 2\ \mu\text{F}$.

The charge on Branch 2 is:
$Q_2 = C_{s2} \times V = 2\ \mu\text{F} \times 18\text{ V} = 36\ \mu\text{C}$.
Since $C_3$ and $C_4$ are in series, the charge on $C_4$ is $36\ \mu\text{C}$.
The electrostatic energy stored in $C_4$ is:
$U_4 = \dfrac{Q_2^2}{2C_4} = \dfrac{(36\ \mu\text{C})^2}{2 \times 6\ \mu\text{F}} = \dfrac{1296}{12}\ \mu\text{J} = 108\ \mu\text{J}$.

The ratio of the energy stored in $C_1$ to that in $C_4$ is:
$\dfrac{U_1}{U_4} = \dfrac{144}{108} = \dfrac{4}{3}$.

Answer: $4 : 3$