Answer
$$2\ \mu \mathrm{W} \text { }
$$
Work Step by Step
We use $$\beta=10 \log \left(I / I_{0}\right)$$ with $$I_{0}=1 \times 10^{-12} \mathrm{W} / \mathrm{m}^{2}$$ and $$I=P / 4 \pi r^{2}$$
We estimate $r \approx 0.3 \mathrm{m}$ and find
$$
P \approx 4 \pi(0.3 \mathrm{m})^{2}\left(1 \times 10^{-12} \mathrm{W} / \mathrm{m}^{2}\right) 10^{6.2}=2 \times 10^{-6} \mathrm{W}=2\ \mu \mathrm{W} \text { }
$$
