Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that the electric field inside a resistor is given by
$$E=\dfrac{\Delta V_R}{L}$$
where $\Delta V_R=IR$
$$E=\dfrac{IR}{L}$$
Hence,
$$\boxed{E_{\rm surface}=\dfrac{IR}{L}}$$
Now we need to find the magnetic field,
$$\oint \vec B\cdot d\vec s=\mu_0I_{\rm through}$$
The wire is cylindrically shaped, so
$$B(2\pi r)=\mu_0I $$
Hence,
$$B=\dfrac{\mu_0I }{2\pi r}$$
Thus,
$$\boxed{B_{\rm surface}=\dfrac{\mu_0I }{2\pi r}}$$
$$\color{blue}{\bf [b]}$$
We know that the Poynting vector is given by
$$\vec S=\dfrac{\vec E\times \vec B}{\mu_0}$$
And its direction is into the curved surface.
Now we can find its magnitude by plugging from the two boxed formula above.
$$S=\dfrac{1}{\color{red}{\bf\not} \mu_0}\left[\dfrac{IR}{L}\times \dfrac{\color{red}{\bf\not} \mu_0I }{2\pi r}\right]$$
$$\boxed{S= \dfrac{I^2R}{2\pi r L} }$$
$$\color{blue}{\bf [c]}$$
Integrating as the author asked,
$$\oint \vec S\cdot d \vec A=\int_{\rm surface}\vec S\cdot d \vec A+2\int_{\rm base}\vec S\cdot d \vec A$$
The cylindrical wire has two bases and an external surface.
$$\oint \vec S\cdot d \vec A=\int_{\rm surface} S d A\cos0^\circ+2\int_{\rm base} S dA \cos 90^\circ$$
$$\oint \vec S\cdot d \vec A=\int_{\rm surface} S d A =S(2\pi r L)$$
Plug $S$ from the last boxed formula above,
$$\oint \vec S\cdot d \vec A= \dfrac{I^2R}{2\pi r L}(2\pi r L)$$
$$\boxed{\oint \vec S\cdot d \vec A= I^2R}$$
This is also the value of the dissipated power by the resistor.
Recalling that the Poynting vector is given by $P/A$ where $P$ is the power and $A$ is the unit area.
