= Solution
Use the <divergence> identity for a <cross product>, followed by <Faraday's law of induction> and the <Ampère-Maxwell equation>:
$$
\nabla\cdot(\mathbf E\times\mathbf H)
=\mathbf H\cdot(\nabla\times\mathbf E)-\mathbf E\cdot(\nabla\times\mathbf H),
$$
$$
\nabla\times\mathbf E=-\mu_0\frac{\partial\mathbf H}{\partial t},\qquad
\nabla\times\mathbf H=\mathbf J+\epsilon_0\frac{\partial\mathbf E}{\partial t}.
$$
With the <electromagnetic energy> density of part (b), these give
$$
\nabla\cdot\mathbf N
=-\mathbf E\cdot\mathbf J-\mu_0\mathbf H\cdot\partial_t\mathbf H-\epsilon_0\mathbf E\cdot\partial_t\mathbf E
=-\mathbf E\cdot\mathbf J-\frac{\partial U}{\partial t}.
$$
Integrate over a fixed volume $V$ and apply the <divergence theorem>, choosing $d\mathbf S$ outward:
$$
\boxed{\oint_S\mathbf N\cdot d\mathbf S
=-\int_V\mathbf E\cdot\mathbf J\,dV-\frac d{dt}\int_VU\,dV.}
$$
This is <Poynting's theorem>. The outward <electromagnetic energy> flux equals minus the <power> transferred to matter and minus the rate at which <electromagnetic energy> accumulates inside. For the dissipative <electrical conductor> in this question, the matter term is <Ohmic heating>.
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