Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know, from Ohm's law, that
$$\varepsilon_{\rm rms}= I_{\rm rms} Z= I_{\rm rms}{\sqrt{(X_L -X_C)^2+R^2}}$$
where $X_L=\omega L=2\pi fL$, and $X_C=1/2\pi fC$
$$\varepsilon_{\rm rms}= I_{\rm rms}\sqrt{\left(2\pi f L-\dfrac{1}{2\pi fC}\right)^2+R^2}$$
Plug the known;
$$\varepsilon_{\rm rms}= (2.5)\sqrt{\left(2\pi (60)(0.10)-\dfrac{1}{2\pi (60)(100\times 10^{-6})}\right)^2+(25)^2}$$
$$\varepsilon_{\rm rms}= \color{red}{\bf 68.5}\;\rm V$$
$$\color{blue}{\bf [b]}$$
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;
$$\phi=\tan^{-1}\left[\dfrac{2\pi (60)(0.10)-\dfrac{1}{2\pi (60)(100\times 10^{-6})}}{(25)}\right]$$
$$\phi=\color{red}{\bf 24.1}^\circ$$
$$\color{blue}{\bf [c]}$$
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;
$$P_R= \dfrac{(68.5)^2\cos(24.1^\circ)}{25}$$
$$P_R= \color{red}{\bf 171}\;\rm W$$
