Answer
(a) $C = 4.35 \times 10^{-12} \,\text{F}$
(b) $V_{ab}$ = 2.3 V
Work Step by Step
(a) The capacitance for a cylindrical shape is given by
$$C = \dfrac{2\pi \epsilon_o}{\ln(r_b/r_a)} L$$
Substitute to get $C$
\begin{align}
C &= \dfrac{2\pi \epsilon_o}{\ln(r_b/r_a)} L \\
&= \dfrac{2\pi (8.85 \times 10^{-12} \,\text{F/m})}{\ln(5.0 \times 10^{-3} \,\text{m}/0.50 \times 10^{-3} \,\text{m})} (0.18 \,\text{m})\\
&= 4.35 \times 10^{-12} \,\text{F}
\end{align}
(b) The capacitance between the two plates occur due to the accumulation of the charges on the plates and the potential is given by
$$V_{ab} = \dfrac{Q}{C} = \dfrac{10 \times 10^{-12} \,\text{F}}{4.35 \times 10^{-12} \,\text{F}} = 2.3 \,\text{V}$$
