Answer
(a) $U = 420 \,\text{J} $
(b) $C = 55.5 \times 10^{-3} \,\text{F} $
Work Step by Step
(a) First, we find the energy use in the flash where the energy is the power times the time
$$E = P \times t = 2.7 \times 10^{5} \,\text{W} \times \dfrac{1}{675} \,\text{s} = 400 \,\text{J}$$
400 J represents 95 % of the energy stored in the capacitor, therefore, we could get the energy stored in the capacitor by
$$U = 400 \,\text{J} + \dfrac{100 \%-95 \%}{100 \%} \times 400 \,\text{J} = \boxed{420 \,\text{J} }$$
(b) We will use the value of the energy stored to get the capacitance $C$
\begin{gather*}
U = \dfrac{1}{2} C V^2 \\
C =\sqrt{\dfrac{2 U}{V^2}} \\ \
C = \dfrac{2\times 400 \,\text{J}}{(120 \,\text{V})^2} \\
\boxed{C = 55.5 \times 10^{-3} \,\text{F} }
\end{gather*}
