Answer
(a) $r_b$ = 0.25 cm
(b) $\lambda$ = 38 nC/m
Work Step by Step
The radius $r_b$ is calculated by
\begin{gather*}
C = \dfrac{2\pi \epsilon_o}{\ln(r_b/r_a)} L\\
\ln(r_b/r_a) = \dfrac{2\pi \epsilon_o}{C} L\\
\dfrac{r_b}{r_a} = e^{\dfrac{2\pi \epsilon_o}{C} L}\\
r_b = r_a e^{\left(\dfrac{2\pi \epsilon_o}{C} L\right)} \tag{1}
\end{gather*}
Substitute with the values of $r_b, L, C$ and $\epsilon_o$ to find $r_a$
\begin{align*}
r_b &= r_a e^{\left(\dfrac{2\pi \epsilon_o}{C} L\right)} \\
&= (0.250 \times 10^{-2} \,\text{m})\, e^{\left(\dfrac{2\pi(8.85 \times 10^{-12} \,\text{F/m})(12 \times 10^{-3}\,\text{m}) }{36.7\times 10^{-12} \,\text{F} } \right)}\\
&= 0.25 \,\text{cm}
\end{align*}
(b) The charge per unit length could be calculated from
\begin{gather*}
\lambda L = CV \\
\lambda = \dfrac{CV}{L} \tag{2}
\end{gather*}
Substitute with the values of $V, C$ and $L$ to find $\lambda$
\begin{align*}
\lambda &= \dfrac{CV}{L} \\
&= \dfrac{(36.7\times 10^{-12} \,\text{F})(125 \,\text{V})}{12 \times 10^{-2}\,\text{m}}\\
&= 38 \,\text{nC/m}
\end{align*}
