Answer
$P = 1.1\times 10^{-10}~W$
Work Step by Step
We can find the photon energy:
$E = \frac{hc}{\lambda}$
$E = \frac{(6.63\times 10^{-34}~J\cdot s)(3.0\times 10^8~m/s)}{500\times 10^{-9}~m}$
$E = 3.978\times 10^{-19}~J$
Since the detector absorbs 80% of the incident light, the number of photons each second in the area of the detector is $5.000~s^{-1}$
We can find the power of the emitter:
$P = I~A$
$P = [\frac{(5.000)(3.978\times 10^{-19}~J)}{2.00\times 10^{-6}~m^2}]~(4\pi)(3.00~m)^2$
$P = 1.1\times 10^{-10}~W$
