Answer
$P_{av}=8.5\times10^{5}W $.
Work Step by Step
For a perfectly absorbing surface, we can relate the radiation pressure to the intensity.
$$p_{rad}=\frac{I}{c}$$
$$I = c (p_{rad})=2.70\times10^3W/m^2$$
At a distance r from the source, the total average power from the star is spread over a spherical surface, of area $4 \pi r^2$. This equals the intensity at that distance.
$$P_{av}=I4\pi r^2=(2.70\times10^3W/m^2)(4 \pi)(5.0m)^2=8.5\times10^{5}W $$