Answer
$\int \vec{S} \cdot d \vec{A}=i^2 R .$
Work Step by Step
The magnitudes of the electric and magnetic fields are $E=V / l=i R / l$ and $B=\mu_0 i / 2 \pi a$, respectively. Thus,
$
S=\frac{E B}{\mu_0}=\frac{1}{\mu_0}\left(\frac{i R}{l}\right)\left(\frac{\mu_0 i}{2 \pi a}\right)=\frac{i^2 R}{2 \pi a l} .
$
Noting that the magnitude of the Poynting vector $S$ is constant, we have
$
\int \vec{S} \cdot d \vec{A}=S A=\left(\frac{i^2 R}{2 \pi a l}\right)(2 \pi a l)=i^2 R .
$
