= Solution
On the curved wire <surface>, the outward normal is $\widehat{\mathbf r}$ and the <Poynting vector> has the constant radial value $N_r(a)=-I^2/(2\sigma\pi^2a^3)$. Its outward <flux> is
$$
\Phi_N=\int_{\rm curved}\mathbf N\cdot d\mathbf S
=N_r(a)(2\pi aL)
=\boxed{-\frac{I^2L}{\sigma\pi a^2}=-P_{\rm loss}.}
$$
The <Poynting vector> is radial, so it has zero normal component on the two end faces in this approximation. This is therefore also the outward <flux> through the complete wire <surface>. Equivalently, positive electromagnetic <power> $P_{\rm loss}$ flows inward.
The fields are steady, so the stored <electromagnetic energy> does not change. <Poynting's theorem> consequently requires the incoming <power> to equal the <Ohmic heating> in part (e), exactly as found. The energy reaches the wire through the surrounding <electromagnetic field>. A local check is $\nabla\cdot\mathbf N=r^{-1}\partial_r(rN_r)=-I^2/(\sigma\pi^2a^4)=-\mathbf E\cdot\mathbf J$.
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