Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
First of all, we need to find $\Delta V_C$ where we know that $\Delta V=-\int Edr $
$$\Delta V_C=-\int_{R_1}^{R_2}Edr\tag 1$$
Now we need to find the electric field inside the coaxial cable which is uniform and radial from the inner wire to the outer one and it is always normal to the surface.
For an infinity-long coaxial cable, the electric field is similar to the electric field of an infinity cylinder that has a linear density of $\lambda$ which is given by
$$E=\dfrac{\lambda}{2\pi \epsilon_0 r }$$
Plug into (1);
$$\Delta V_C=-\dfrac{\lambda}{2\pi \epsilon_0 }\int_{R_1}^{R_2}\dfrac{1}{r} dr$$
$$\Delta V_C=-\dfrac{\lambda}{2\pi \epsilon_0 }\int_{R_1}^{R_2} \ln r\bigg|_{R_1}^{R_2}$$
$$\Delta V_C=-\dfrac{\lambda}{2\pi \epsilon_0 } \ln \left[ \dfrac{R_2}{R_1}\right]\tag 1$$
In the figure below, we draw a small part of our coaxial cable. Let's assume that this part has a length of $L$ and a net charge of $Q$ where $$Q=\lambda L\tag 2$$
And we know, for a parallel-plate capacitor, that the capacitance is given by
$$C=\dfrac{Q}{\Delta V}$$
Plug from (1) and neglect the negative sign, and (2);
$$C=\dfrac{\lambda L}{\dfrac{\lambda}{2\pi \epsilon_0 } \ln \left[ \dfrac{R_2}{R_1}\right]}$$
Solving for $C/L$ which is the capacitance per meter,
$$\boxed{\dfrac{C}{L}=\dfrac{2\pi \epsilon_0 }{ \ln \left[ \dfrac{R_2}{R_1}\right]}}$$
$$\color{blue}{\bf [b]}$$
Plug the known into the boxed formula above,
$$ \dfrac{C}{L}=\dfrac{2\pi(8.85\times 10^{-12}) }{ \ln \left[ \dfrac{3}{0.5}\right]} $$
$$ \dfrac{C}{L}=\color{red}{\bf 31}\;\rm pF/m$$
