An 18 µF capacitor is joined in series with a parallel pair of 3.0 µF and 6.0 µF…

Step 1. The pair. $3.0\ \mu\text{F} + 6.0\ \mu\text{F} = 9.0\ \mu\text{F}$.

Step 2. The series. $C = \dfrac{18\ \mu\text{F} \times 9.0\ \mu\text{F}}{18\ \mu\text{F} + 9.0\ \mu\text{F}} = 6.0\ \mu\text{F}$.

Step 3. The charge. $Q = 6.0\ \mu\text{F} \times 36\ \text{V} = 216\ \mu\text{C}$. This is on the $18\ \mu\text{F}$ capacitor.

Step 4. The voltages. Across the $18\ \mu\text{F}$ capacitor: $\dfrac{216\ \mu\text{C}}{18\ \mu\text{F}} = 12\ \text{V}$. Across the pair: $36\ \text{V} - 12\ \text{V} = 24\ \text{V}$.

Step 5. The charge on the $6.0\ \mu\text{F}$ capacitor. $6.0\ \mu\text{F} \times 24\ \text{V} = 144\ \mu\text{C}$. The $3.0\ \mu\text{F}$ capacitor has $72\ \mu\text{C}$.