Answer
(a) $0.10KW$
(b) $0.55H$
Work Step by Step
(a) We can find the required power as
$P_{avg}=\frac{V_{rms}^2 }{R}$
We plug in the known values to obtain:
$P_{avg}=\frac{(110)^2}{120}$
$P_{avg}=100W=0.10KW$
(b) We know that
$L=\frac{3R^2}{\omega^2}$
$L=\sqrt 3\frac{R}{\omega}$
We plug in the known values to obtain:
$L=\sqrt3 \frac{120}{2\pi (60)}$
$L=0.55H$
