= Solution
Take the <electric current> along $+z$, and use <cylindrical coordinates>. A <uniform current in a straight homogeneous wire> has
$$
\mathbf J=\frac I{\pi a^2}\widehat{\mathbf z}.
$$
The local <Ohm's law> $\mathbf J=\sigma\mathbf E$ gives the longitudinal <electric field>
$$
\mathbf E=\frac I{\sigma\pi a^2}\widehat{\mathbf z}.
$$
For a circular <Amperian loop> of radius $r\le a$, the enclosed <electric current> is $I_{\rm enc}=Ir^2/a^2$. The <magnetostatic Ampère law> gives $2\pi rH_\phi=I_{\rm enc}$, hence
$$
\mathbf H=\frac{Ir}{2\pi a^2}\widehat{\boldsymbol\phi}.
$$
Since $\widehat{\mathbf z}\times\widehat{\boldsymbol\phi}=-\widehat{\mathbf r}$, the <Poynting vector> is
$$
\boxed{\mathbf N(r)=-\frac{I^2r}{2\sigma\pi^2a^4}\widehat{\mathbf r},\qquad0\le r\le a.}
$$
It is zero at the centre and points radially inward elsewhere. Its units are $\mathrm{W\,m^{-2}}$. These are the interior fields in the long-wire approximation, neglecting end and contact effects.
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