Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that the $\varepsilon_{\rm rms}=\varepsilon_0/\sqrt{2}$, so
$$\varepsilon_0=\sqrt{2}\varepsilon_{\rm rms}=\sqrt{2}(120) $$
$$\varepsilon_0= \color{red}{\bf 170}\;\rm V$$
$$\color{blue}{\bf [b]}$$
The phase angle is given by
$$\phi=\tan^{-1}\left[\dfrac{X_L-X_C}{R}\right]$$
where $X_L=\omega L=2\pi fL$, and $X_C=1/2\pi fC$
$$\phi=\tan^{-1}\left[\dfrac{2\pi fL-\dfrac{1}{2\pi fC}}{R}\right]$$
Plug the given;
$$\phi=\tan^{-1}\left[\dfrac{2\pi (60)(0.15)-\dfrac{1}{2\pi (60)(30\times 10^{-6})}}{(100)}\right]$$
$$\phi=\color{red}{\bf -17.7}^\circ$$
$$\color{blue}{\bf [c]}$$
The average power loss is the power dissipated through the resistor which is given by
$$P_R=I_{\rm rms}\varepsilon_{\rm rms}\cos\phi$$
where $I_{\rm rms}=\varepsilon_{\rm rms}/R$,
$$P_R= \dfrac{\varepsilon^2_{\rm rms}\cos\phi}{R}$$
Plug the known;
$$P_R= \dfrac{(120)^2\cos(-17.7^\circ)}{100}$$
$$P_R= \color{red}{\bf 137}\;\rm W$$
