= Solution
<Gauss's law> in electrostatics states
$$
\oint_{\partial V}E\mathbin{\cdot}dS
=\frac{Q_{\rm enclosed}}{\varepsilon_0}.
$$
Cylindrical symmetry and a coaxial Gaussian cylinder give
$$
E(r)=
\begin{cases}
0,&0<r<a,\\[2pt]
\dfrac{Q}{2\pi\varepsilon_0r}\,e_r,&a<r<b,\\[6pt]
0,&r>b.
\end{cases}
$$
The field vanishes inside each perfect conductor, and outside the cable because the total enclosed charge per unit length is zero.
Choose the outer conductor's potential to be zero. Since $E=-\nabla V$,
$$
V(r)=
\begin{cases}
\dfrac{Q}{2\pi\varepsilon_0}\log(b/a),&0<r\leq a,\\[6pt]
\dfrac{Q}{2\pi\varepsilon_0}\log(b/r),&a<r<b,\\[6pt]
0,&r\geq b.
\end{cases}
$$
The <capacitance per unit length> is therefore
$$
\boxed{C=\frac{Q}{V(a)-V(b)}
=\frac{2\pi\varepsilon_0}{\log(b/a)}}.
$$
The <electrostatic energy> per unit length is
$$
U=\frac{\varepsilon_0}{2}
\int_a^b|E|^2\,2\pi r\,dr
=\boxed{\frac{Q^2}{4\pi\varepsilon_0}\log\frac ba}.
$$
Substitution of $C$ verifies $U=Q^2/(2C)$.
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