Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that the average power for an $RLC$ circuit is given by
$$P_{\rm source}=I_{\rm rms}\varepsilon_{\rm rms}\cos\phi\tag 1$$
where
$$I=\dfrac{\varepsilon_{\rm rms}}{Z}\tag 2$$
And from the geometry of the figure below, we can see that
$$\cos\phi=\dfrac{R}{Z}\tag 3$$
Plug (3), and (2) into (1):
$$P_{\rm source}=\dfrac{\varepsilon_{\rm rms}}{Z} \dfrac{R}{Z}\varepsilon_{\rm rms} $$
$$P_{\rm source}=\dfrac{\varepsilon_{\rm rms}^2R}{Z^2} $$
Recalling that $Z^2=(X_L-X_C)^2+R^2 $
$$P_{\rm source}=\dfrac{\varepsilon_{\rm rms}^2R}{(X_L-X_C)^2+R^2} $$
where $X_L=\omega L$, and $X_C=1/\omega C$
$$P_{\rm source}=\dfrac{\varepsilon_{\rm rms}^2R}{(\omega L-\frac{1}{\omega C})^2+R^2} $$
$$P_{\rm source}=\dfrac{\varepsilon_{\rm rms}^2R}{\frac{L^2}{\omega^2}(\omega^2 -\frac{1}{LC})^2+R^2} $$
where $1/LC=\omega_0^2$;
$$\boxed{P_{\rm avg}=\dfrac{\varepsilon_{\rm rms}^2\omega^2R}{ L^2 (\omega^2 -\omega_0^2)^2+\omega^2R^2} }$$
$$\color{blue}{\bf [b]}$$
Maximizing the energy dissipation means
$$\dfrac{dP_{\rm avg}}{d\omega}=0$$
$$0=\dfrac{d}{d\omega} \dfrac{\varepsilon_{\rm rms}^2\omega^2R}{ L^2 (\omega^2 -\omega_0^2)^2+R^2} $$
$$ \dfrac{2\varepsilon_{\rm rms}^2\omega R [L^2 (\omega^2 -\omega_0^2)^2+\omega^2R^2]-[\varepsilon_{\rm rms}^2\omega^2 R] [4L^2\omega (\omega^2 -\omega_0^2) +2\omega R^2 ]}{ [L^2 (\omega^2 -\omega_0^2)^2 +\omega^2R^2]^2} =0$$
Hence,
$$2\varepsilon_{\rm rms}^2\omega R [L^2 (\omega^2 -\omega_0^2)^2+\omega^2R^2]-[\varepsilon_{\rm rms}^2\omega^2 R] [4L^2\omega (\omega^2 -\omega_0^2)+2\omega R^2 ]=0$$
$$2 [L^2 (\omega^2 -\omega_0^2)^2+\omega^2R^2]- \omega [4L^2\omega (\omega^2 -\omega_0^2) +2\omega R^2 ]=0$$
$$ 2L^2 (\omega^2 -\omega_0^2)^2+2\omega^2R^2 = 4L^2\omega^2 (\omega^2 -\omega_0^2) +2\omega^2 R^2 $$
$$ 2L^2 (\omega^4 -2\omega^2\omega_0^2+\omega_0^4) +2\omega^2R^2 = 4L^2\omega^4 -4L^2\omega^2\omega_0^2+2\omega^2 R^2 $$
$$ 2L^2 \omega^4 -\color{magenta}{ 4L^2\omega^2\omega_0^2}+2L^2\omega_0^4 +\color{blue}{ 2\omega^2R^2} = 4L^2\omega^4 -\color{magenta}{ 4L^2\omega^2\omega_0^2} +\color{blue}{ 2\omega^2R^2} $$
$$ 2L^2 \omega^4 +2L^2\omega_0^4 = 4L^2\omega^4 $$
$$ \omega^4 + \omega_0^4 =2\omega^4 $$
$$\omega^4=\omega_0^4$$
Therefore,
$$ \boxed{ \omega =\pm\;\omega_0 }$$
