Solution

The rate at which photons are absorbed by the detector is related to the rate of photon emission by the light source via
$$R_{\mathrm{abs}}=(0.80) \dfrac{A_{\mathrm{abs}}}{4 \pi r^{2}} R_{\mathrm{emit}}\\$$
Given that $$A_{\text {bs }}=2.00 \times 10^{-6} \mathrm{m}^{2}$$ and $$r=3.00 \mathrm{m},$$ with $$R_{\text {abs }}=4.000$$ photons/s, we find the rate at which photons are emitted to be
$$R_{\text {emit }}=\dfrac{4 \pi r^{2}}{(0.80) A_{\text {abs }}} R_{\text {abs }}=\dfrac{4 \pi(3.00 \mathrm{m})^{2}}{(0.80)\left(2.00 \times 10^{-6} \mathrm{m}^{2}\right)}(4.000 \text { photons } / \mathrm{s})=2.83 \times 10^{8} \text { photons } / \mathrm{s}\\$$
since the energy of each emitted photon is
$$E_{\mathrm{ph}}=\dfrac{h c}{\lambda}=\dfrac{1240 \mathrm{eV} \cdot \mathrm{nm}}{500 \mathrm{nm}}=2.48 \mathrm{eV}\\$$
the power output of source is
$$P_{\mathrm{emit}}=R_{\mathrm{emit}} E_{\mathrm{ph}}=\left(2.83 \times 10^{8} \mathrm{photons} / \mathrm{s}\right)(2.48 \mathrm{eV})=7.0 \times 10^{8} \mathrm{eV} / \mathrm{s}=1.1 \times 10^{-10} \mathrm{W}\\$$