Due to flow of current through the cable, shown in Fig (a), magnetic field is established in the space between cylinders. Energy is stored in this space due to magnetic field. 

Since energy stored per unit volume in magnetic field of induction $B$ is equal to $B^2/2{\mu }_0$, therefore magnetic induction in the space must be known. But magnetic field in the space is not uniform.

Hence consider a thin cylindrical coaxial shell in the space. Let radius of the shell be $x$ and radial thickness $dx$ as shown in Fig. (b)

According to Ampere’s circuital law, magnetic induction $B$ is given by $B2\pi x={\mu }_{0}I$

Or, $B=\frac{{\mu }_{0}I}{2\pi x}$

Therefore, Energy stored per unit length of the shell considered is 

$dU=\frac{B^2}{2{\mu }_0}\left(2\pi xdx\right)=\frac{{\mu }_0{I}^2}{4\pi x}dx$

Therefore Energy stored per unit length of the cable,

$U=\int dU={\int }_{x=R}^{x=2R}\frac{{\mu }_0{I}^2}{4\pi x}dx=\frac{{\mu }_0{I}^2}{4\pi }{\log }_{e}2$