Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
The inner wire creates a circular magnetic field around it that fills the space between the inner and the outer conductors (wires).
Let's assume that the triangle between the two conductors is perpendicular to the field and has a length of $l$.
Hence, the small flux exerted on a very this part of this triangle, that has a width of $dr$, is given by
$$d\Phi=BA\cos0^\circ=BA$$
where $A=ldr$
$$d\Phi= Bldr$$
where $B$ is from the inner wire which is given by $\mu_0I/2\pi r$
$$d\Phi= \dfrac{\mu_0I}{2\pi r}ldr$$
$$d\Phi= \dfrac{\mu_0Il}{2\pi r}dr$$
Integrating,
$$\int_0^\Phi d\Phi= \dfrac{\mu_0Il}{2\pi }\int_{r_1}^{r_2} \dfrac{dr}{r} $$
$$ \Phi= \dfrac{\mu_0Il}{2\pi } \ln\left[\dfrac{ r_2}{r_1} \right]\tag 1$$
We know that
$$L=\dfrac{\Phi}{I}$$
and hence the inductance per meter is given by
$$\dfrac{L}{l}=\dfrac{\Phi}{Il}$$
Plug from (1),
$$\dfrac{L}{l}=\dfrac{ \mu_0Il}{2\pi Il} \ln\left[\dfrac{ r_2}{r_1} \right]$$
$$\boxed{\dfrac{L}{l}=\dfrac{ \mu_0 }{2\pi } \ln\left[\dfrac{ r_2}{r_1} \right]}$$
$$\color{blue}{\bf [b]}$$
Plug the known into the boxed formula above,
$$ \dfrac{L}{l}=\dfrac{ 4\pi\times 10^{-7}}{2\pi } \ln\left[\dfrac{ 3}{0.5} \right] $$
$$ \dfrac{L}{l}=\color{red}{\bf 3.58\times 10^{-7}}\;\rm H/m$$
