Answer
$P = 8.85\times 10^{-10}~W$
Work Step by Step
We can find the amplitude of the resultant wave when the phase difference is $0.40 \pi~rad$:
$s_m' = 2~s_m~cos~\frac{\phi}{2}$
$s_m' = (2)~(12.0~nm)~(cos~\frac{0.40~\pi}{2})$
$s_m' = 19.416~nm$
We can find the average power:
$P = I~A$
$P = \frac{1}{2}\rho~v~\omega^2~(s_m')^2~\pi~r^2$
$P = (\frac{1}{2})(1.21~kg/m^3)(343~m/s)~(3000~rad/s)^2~(19.416\times 10^{-9}~m)^2~(\pi)~(2.00\times 10^{-2}~m)^2$
$P = 8.85\times 10^{-10}~W$
