Answer
(a) $U = 1.57 \times 10^{-4} \,\text{J}$
(b) $U = 72 \times 10^{-6}\,\text{J}$
Work Step by Step
(a) We will find the equivalent capacitance of the network first to get the energy storage
The capacitance in series for two capacitors is given by
\begin{align*}
C^1 = \dfrac{C_1 C_2}{C_1 +C_2} = \dfrac{( 6.20 \,\text{uF}) (11.80 \,\text{uF})}{6.20 \,\text{uF} + 11.80 \,\text{uF}} = 4.06 \,\text{uF}
\end{align*}
In parallel, the equivalent capacitance of both capacitors is given by
$$C^2 = 4.06 \,\text{uF} + 3.50 \,\text{uF} = 7.56 \,\text{uF} $$
Also, in series, the capacitance is given by
$$\dfrac{1}{C_{eq} }= \dfrac{1}{7.56 \,\text{uF} } + \dfrac{1}{8.60 \,\text{uF} } + \dfrac{1}{4.80 \,\text{uF} } $$
$$C_{eq} = 2.19 \,\text{uF}$$
The stored energy in the network is given by:
$$U_1 = \dfrac{1}{2} C_{eq}V^2 = \dfrac{1}{2} (2.19 \times 10^{-6} \,\text{F})(12 \,\text{V})^2 = \boxed{1.57 \times 10^{-4} \,\text{J}}$$
(b) We could get the charge in the capacitor $4.80 \,\text{uF}$ where the charge on it is the same for the total network and it is calculated by
$$Q = C V = (2.19 \times 10^{-6}\,\text{F}) (12 \,\text{V}) = 26.28 \times 10^{-6}\,\text{C} $$
Hence, the energy stored in $4.80 \,\text{uF}$ is given by
$$U= \dfrac{Q^2}{2C} = \dfrac{(26.28 \times 10^{-6}\,\text{C})^2}{2(4.80 \times 10^{-6}\,\text{F})} = \boxed{72 \times 10^{-6}\,\text{J}}$$
