A circuit is connected to a 24 V battery. The circuit consists of...

The equivalent capacitance of $C_1$ and $C_2$ in series is $C_{12} = \dfrac{6 \times 3}{6 + 3} = 2\ \mu\text{F}$.

This combination is in parallel with $C_3 = 2\ \mu\text{F}$, so the equivalent capacitance of the parallel block is $C_{123} = 2 + 2 = 4\ \mu\text{F}$.

The parallel block is in series with $C_4 = 4\ \mu\text{F}$. The total equivalent capacitance of the circuit is $C_{eq} = \dfrac{4 \times 4}{4 + 4} = 2\ \mu\text{F}$. This matches (A) with (II).

The total charge drawn from the battery is $Q_{total} = C_{eq} \times V = 2\ \mu\text{F} \times 24\text{ V} = 48\ \mu\text{C}$.
Since $C_4$ is in series with the rest of the circuit, the charge on $C_4$ is the total charge, $48\ \mu\text{C}$. This matches (B) with (III).

The potential difference across $C_4$ is $V_4 = \dfrac{Q_4}{C_4} = \dfrac{48\ \mu\text{C}}{4\ \mu\text{F}} = 12\text{ V}$.
The potential difference across the parallel block is $V_{123} = 24\text{ V} - 12\text{ V} = 12\text{ V}$.

The charge on $C_3$ is $Q_3 = C_3 \times V_{123} = 2\ \mu\text{F} \times 12\text{ V} = 24\ \mu\text{C}$. This matches (D) with (IV).

The charge on the series branch containing $C_1$ and $C_2$ is $Q_{12} = C_{12} \times V_{123} = 2\ \mu\text{F} \times 12\text{ V} = 24\ \mu\text{C}$.
Since $C_1$ and $C_2$ are in series, the charge on $C_1$ is $24\ \mu\text{C}$.
The potential difference across $C_1$ is $V_1 = \dfrac{Q_1}{C_1} = \dfrac{24\ \mu\text{C}}{6\ \mu\text{F}} = 4\text{ V}$. This matches (C) with (I).

Answer: (A)-(II), (B)-(III), (C)-(I), (D)-(IV)