A capacitor C_1 = 6 is connected in series with a parallel...

First, find the equivalent capacitance of the parallel combination of $C_2$ and $C_3$:
$C_p = C_2 + C_3 = 1 \mu\text{F} + 2 \mu\text{F} = 3 \mu\text{F}$

The total equivalent capacitance of the circuit is:
$C_{eq} = \dfrac{C_1 \times C_p}{C_1 + C_p} = \dfrac{6 \times 3}{6 + 3} = 2 \mu\text{F}$

The total charge drawn from the source is:
$Q_{total} = C_{eq}V = 2V$

Since $C_1$ is in series with the rest of the circuit, the entire total charge flows through it:
$Q_1 = Q_{total} = 2V$

This total charge $Q_{total}$ then splits between the parallel branches $C_2$ and $C_3$ in direct proportion to their capacitances. The charge on $C_2$ is:
$Q_2 = \left(\dfrac{C_2}{C_2 + C_3}\right) Q_{total} = \left(\dfrac{1}{1 + 2}\right) (2V) = \dfrac{2V}{3}$

The ratio of the charge on $C_1$ to the charge on $C_2$ is:
$\dfrac{Q_1}{Q_2} = \dfrac{2V}{\dfrac{2V}{3}} = 3$

Answer: $3$