= Solution
Since $\mathbf e_\rho\times\mathbf e_\varphi=\mathbf e_z$, the <Poynting vector> is axial. The <magnetic permeability> cancels from $\mathbf E\times\mathbf B/\mu$ in each material layer:
$$
\boxed{\boldsymbol{\mathcal P}(\rho)=\begin{cases}
0,&\rho<a\ \text{or}\ \rho>b,\\[2pt]
\displaystyle\frac{VI\epsilon'}{2\pi\Delta\rho^2}\mathbf e_z,&a<\rho<c,\\[2pt]
\displaystyle\frac{VI\epsilon}{2\pi\Delta\rho^2}\mathbf e_z,&c<\rho<b.
\end{cases}}
$$
The signed <Poynting flux> through the plane with <unit normal> $+\mathbf e_z$ is its <surface integral> with area element $\rho\,d\rho\,d\varphi$:
$$
\begin{aligned}
P&=2\pi\int_a^b\mathcal P_z(\rho)\rho\,d\rho\\
&=\frac{VI}{\Delta}\left[\epsilon'\ln(c/a)+\epsilon\ln(b/c)\right].
\end{aligned}
$$
Thus
$$
\boxed{P=VI.}
$$
The transmitted <electromagnetic energy> per unit time is the circuit's terminal power. For $VI>0$, the <Poynting flux> points in the direction of the inner <electric current>. The <electromagnetic energy> travels through the insulating gap and reaches a load at the end of the <coaxial cable>. The radii and material constants redistribute the <electromagnetic energy density> and the local <Poynting vector>, while their integrated <Poynting flux> depends only on the terminal <electric potential difference> and <electric current>.
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