= Solution
The <Poynting vector> in the annular region is
$$
\boxed{\mathbf S=\frac1{\mu_0}\mathbf E\times\mathbf B
=\frac{VI}{2\pi r^2\ln(b/a)}\mathbf e_z.}
$$
It points from the battery towards the resistor with the preceding current convention. Integrating through an annular cross-section gives
$$
\mathcal P=\int_a^bS_z\,2\pi r\,dr
=\frac{VI}{\ln(b/a)}\int_a^b\frac{dr}{r}
=\boxed{VI=V^2/R.}
$$
The energy is transported by the electromagnetic field in the space between the <ideal> conductors; the integration recovers the power dissipated at the terminating resistor.
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