Answer
a. 40.0 W.
b. $I_{rms}=0.167A$.
c. $R=720\Omega$.
Work Step by Step
a. The maximum instantaneous power is twice the average power, so it is 40.0 W.
b. $I_{rms}=\frac{P_{av}}{V_{rms}}=\frac{20.0W}{120V}=0.167A$.
c. $R=\frac{P_{av}}{I_{rms}^2}=\frac{20.0W}{(0.167A)^2}=720\Omega$.
