Answer
$4.16W$
Work Step by Step
We know that
$Z=\sqrt{R^2+(\omega C)^{-2}}$
We plug in the known values to obtain:
$Z=\sqrt{(3.30K\Omega)^2+(2\pi(60.0)(2.75\mu F))^{-2}}=3.44K\Omega$
Now $I_{rms}=\frac{V_{rms}}{Z}=\frac{122}{3.44K\Omega}=35.5mA$
We can find the average power consumed as
$P_{avg}=I_{rms}^2R$
We plug in the known values to obtain:
$P_{avg}=(0.0355)^2(3.30K\Omega)=4.16W$