Answer
(a) $C = 3.47 \mathrm{~\mu F}$
(b) $Q = 174 \mathrm{~\mu C}$
(c) $Q = 174 \mathrm{~\mu C}$
Work Step by Step
(a) The two parallel capacitors $5 \mathrm{~\mu F}$ and $8 \mathrm{~\mu F}$ have an equivalent capacitance by
$$C_{eq} = 5 \mathrm{~\mu F} + 8 \mathrm{~\mu F} = 13 \mathrm{~\mu F}$$
Also, in series, the three capacitances $9 \mathrm{~\mu F}, 10 \mathrm{~\mu F}$ and $C_{eq}$ have an equivalent capacitance given by
\begin{align*}
\dfrac{1}{C} = \dfrac{1}{9 \mathrm{~\mu F}} + \dfrac{1}{10 \mathrm{~\mu F}} + \dfrac{1}{13 \mathrm{~\mu F}} = 0.288
\end{align*}
Take the reciprocal
$$\boxed{C = 3.47 \mathrm{~\mu F}}$$
(b) The charge of the combination is calculated by
$$Q = C V_{ab} =(3.47 \mathrm{~\mu F}) (50 \,\text{V}) = \boxed{174 \mathrm{~\mu C}}$$
(c) In series, each capacitor has the same charge of the total combination. Hence, $9 \mathrm{~\mu F}$ and $10 \mathrm{~\mu F}$ the same charge of the combination
$$\boxed{Q = 174 \mathrm{~\mu C}}$$
