Answer
$P = 1.7\times 10^6~W$
Work Step by Step
In part (a), we found the intensity at a distance of $40~m$ to be $83~W/m^2$
We can find the power of a point source providing that intensity at that distance:
$I = \frac{P}{A}$
$P = I~A$
$P = I~4\pi~r^2$
$P = (83~W/m^2)(4\pi)(40~m)^2$
$P = 1.7\times 10^6~W$
