Step 1: Understanding the Concept:
In a series-parallel circuit, the total charge $Q_1$ leaves $C_1$ and splits between $C_2$ and $C_3$. In parallel, charge divides in direct proportion to capacitance: $Q \propto C$.
Step 2: Formula Application:
$Q_1 = Q_2 + Q_3$.
$Q_3 = Q_1 \left( \frac{C_3}{C_2 + C_3} \right)$.
Step 3: Explanation:
Without the specific values of $C_2$ and $C_3$ from your image, we use the logic: $\frac{Q_3}{Q_1} = \frac{C_3}{C_2 + C_3}$.
For example, if $C_3 = 3\mu F$ and $C_2 = 2\mu F$, the ratio would be $3/5 = 0.6$. Based on typical exam diagrams for these options, $Q_3/Q_1$ usually involves the division of total charge.
Step 4: Final Answer:
Please verify the diagram values; the ratio is determined by $\frac{C_3}{C_2 + C_3}$.