Answer
In parallel $Q = 7.2 \times 10^{-3}\,\text{C}$ and in series $Q = 0.654 \times 10^{-3}\,\text{C} $
Work Step by Step
We have two circuits, first when the capacitors are in parallel where, in parallel, the equivalent capacitance is given by
$$C_{eq} = 10 \mathrm{~\mu F} + 20 \mathrm{~\mu F} +30 \mathrm{~\mu F} = 60 \mathrm{~\mu F} $$
We could get the charge and it is calculated by
$$Q = C_{eq} V = (60 \times 10^{-6}\,\text{F}) (120 \,\text{V}) = \boxed{7.2 \times 10^{-3}\,\text{C}} $$
Second, when the capacitors are in series where, in series, the equivalent capacitance is given by
$$\dfrac{1}{C_{eq} }= \dfrac{1}{10 \mathrm{~\mu F} } + \dfrac{1}{20 \mathrm{~\mu F}} + \dfrac{1}{30 \mathrm{~\mu F} } $$
$$C_{eq} = 5.45\mathrm{~\mu F}$$
We could get the charge and it is calculated by
$$Q = C_{eq} V = (5.45 \times 10^{-6}\,\text{F}) (120 \,\text{V}) = \boxed{0.654 \times 10^{-3}\,\text{C}} $$
