Answer
The average rate of power is $~~4~P_{avg,1}$
Work Step by Step
The average rate of power is proportional to $A^2$ where $A$ is the amplitude.
If $\phi = 0$, the waves are in phase, so the interference is fully constructive.
The amplitude of the resultant wave is the sum of the two amplitudes.
If $\phi = 0.2$ a wavelength, the interference is intermediate but closer to fully constructive.
The amplitude of the resultant wave is greater than the amplitude of one wave but less than the sum of the amplitudes of both waves.
If $\phi = 0.5$ a wavelength, the waves are out of phase by half a wavelength, so the interference is fully destructive.
The amplitude of the resultant wave is the difference of the two amplitudes. In this case, the amplitude of the resultant wave is zero.
We can rank those choices of $\phi$ according to the average rate at which the waves will transport energy:
$0 ~wavelength \gt 0.2 ~wavelength \gt 0.5 ~wavelength$
If $\phi = 0$, then the average rate of power is proportional to $(2A)^2$ where $A$ is the amplitude of the original wave.
Therefore, the average rate of power is $~~4~P_{avg,1}$
