Answer
(a) $Q = 1.6 \times 10^{-9} \,\text{C}$
(b) $(r_b/r_a) = 8$
Work Step by Step
(a) We use the value of the energy stored to get the charge in the form
\begin{gather*}
U = \dfrac{1}{2} QV \\
Q = 2 \dfrac{U}{V}\\
Q = 2 \dfrac{3.2 \times 10^{-9} \,\text{J}}{4 \,\text{V}}\\
\boxed{Q = 1.6 \times 10^{-9} \,\text{C}}
\end{gather*}
(b) The capacitance per unit length $C/L$ for a cylindrical shape is given by
$$C/L = \dfrac{2\pi \epsilon_o}{\ln(r_b/r_a)} $$
\begin{gather*}
C/L = \dfrac{2\pi \epsilon_o}{\ln(r_b/r_a)} \\
\ln(r_b/r_a) = \dfrac{2\pi \epsilon_o}{C/L} \tag{take exp} \\
(r_b/r_a) = e^{\left(\dfrac{2\pi \epsilon_o}{C/L} \right)} \tag{1}\\
\end{gather*}
We want to find $C$ before calculate $(r_b/r_a) $ where $C$ is calcauted by
$$C = \dfrac{Q}{V} = \dfrac{1.6 \times 10^{-9} \,\text{C}}{4 \,\text{V}} = 4 \times 10^{-10} \,\text{F}$$
Substitute in equation (1) to get $(r_b/r_a)$
\begin{align*}
(r_b/r_a) &= e^{\left(\dfrac{2\pi \epsilon_o}{C/L} \right)} \\
&= e^{\left(\dfrac{2\pi(8.85 \times 10^{-12} \,\text{F/m})}{(4 \times 10^{-10} \,\text{F})/(15 \,\text{m})} \right)}\\
&= 8
\end{align*}
