Answer
a) ${\bf 4.44\times 10^{-4}}\;\rm W$
b) ${\bf 1.0}\;\rm W$
c) ${\bf 4.44\times 10^{-4}}\;\rm W$
Work Step by Step
We know that the power supplied to the circuit is given by
$$P_{\rm source}=I_{\rm rms}\varepsilon_{\rm rms}\cos\phi\tag 1$$
where
$$I_{\rm rms}=\dfrac{\varepsilon_{\rm rms}}{Z}$$
$$I_{\rm rms}=\dfrac{\varepsilon_{\rm rms}}{\sqrt{(X_L-X_C)^2+R^2}}$$
where $X_L=\omega L$, and $X_C=1/\omega C$
$$I_{\rm rms}=\dfrac{\varepsilon_{\rm rms}}{\sqrt{\left(\omega L-\dfrac{1}{\omega C}\right)^2+R^2}} $$
Plug into (1),
$$P_{\rm source}=\dfrac{\varepsilon_{\rm rms}^2\cos\phi}{\sqrt{\left(\omega L-\dfrac{1}{\omega C}\right)^2+R^2}}\tag 2$$
where $\phi$ is given by
$$\phi=\tan^{-1}\left[\dfrac{X_L-X_C}{R}\right]$$
where $X_L=\omega L$, and $X_C=1/\omega C$
$$\phi=\tan^{-1}\left[\dfrac{\omega L-\dfrac{1}{\omega C}}{R}\right]\tag 3$$
We are given the angular frequency in terms of the resonance frequency where the resonance frequency is given by $\omega_0=1/\sqrt{LC}$
$$\color{blue}{\bf [a]}$$
$\bullet$ When $\omega=\dfrac{1}{2}\omega_0 =\dfrac{1}{2\sqrt{LC}}$,
Plug into (3),
$$\phi=\tan^{-1}\left[\dfrac{\dfrac{1}{2\sqrt{LC}} L-\dfrac{1}{\dfrac{1}{2\sqrt{LC}}C}}{R}\right] $$
$$\phi=\tan^{-1}\left[\dfrac{\sqrt{\dfrac{L}{4 C}} -\dfrac{1}{\sqrt{\dfrac{C}{4L}}}}{R}\right] $$
Plug the known;
$$\phi=\tan^{-1}\left[\dfrac{\sqrt{\dfrac{(0.01)}{4 (1\times 10^{-9})}} -\dfrac{1}{\sqrt{\dfrac{ (1\times 10^{-9})}{4(0.01)}}}}{(100)}\right]=\bf -88.8^\circ $$
Plug into (2),
$$P_{\rm source}=\dfrac{\varepsilon_{\rm rms}^2\cos(-88.8^\circ)}{\sqrt{\left(\sqrt{\dfrac{L}{4 C}} -\dfrac{1}{\sqrt{\dfrac{C}{4L}}}\right)^2+R^2}} $$
Plug the known;
$$P_{\rm source}=\dfrac{(10)^2\cos(-88.8^\circ)}{\sqrt{\left(\sqrt{\dfrac{(0.01)}{4 (1\times 10^{-9})}} -\dfrac{1}{\sqrt{\dfrac{ (1\times 10^{-9})}{4(0.01)}}}\right)^2+100^2}} $$
$$P_{\rm source}=\color{red}{\bf 4.44\times 10^{-4}}\;\rm W$$
$$\color{blue}{\bf [b]}$$
$\bullet$ When $\omega= \omega_0 =\dfrac{1}{ \sqrt{LC}}$,
Plug into (3),
$$\phi=\tan^{-1}\left[\dfrac{\dfrac{1}{ \sqrt{LC}} L-\dfrac{1}{\dfrac{1}{ \sqrt{LC}}C}}{R}\right] $$
$$\phi=\tan^{-1}\left[\dfrac{\sqrt{\dfrac{L}{ C}} -\dfrac{1}{\sqrt{\dfrac{C}{L}}}}{R}\right] $$
Plug the known;
$$\phi=\tan^{-1}\left[\dfrac{\sqrt{\dfrac{(0.01)}{ (1\times 10^{-9})}} -\dfrac{1}{\sqrt{\dfrac{ (1\times 10^{-9})}{ (0.01)}}}}{(100)}\right]=\bf 0^\circ $$
Plug into (2),
$$P_{\rm source}=\dfrac{\varepsilon_{\rm rms}^2\cos(0^\circ)}{\sqrt{\left(\sqrt{\dfrac{L}{4 C}} -\dfrac{1}{\sqrt{\dfrac{C}{4L}}}\right)^2+R^2}} $$
Plug the known;
$$P_{\rm source}=\dfrac{(10)^2 }{\sqrt{\left(\sqrt{\dfrac{(0.01)}{ (1\times 10^{-9})}} -\dfrac{1}{\sqrt{\dfrac{ (1\times 10^{-9})}{ (0.01)}}}\right)^2+100^2}} $$
$$P_{\rm source}=\color{red}{\bf 1.0}\;\rm W$$
$$\color{blue}{\bf [c]}$$
$\bullet$ When $\omega= 2\omega_0 =\dfrac{2}{ \sqrt{LC}}$,
Plug into (3),
$$\phi=\tan^{-1}\left[\dfrac{\dfrac{2}{ \sqrt{LC}} L-\dfrac{1}{\dfrac{2}{ \sqrt{LC}}C}}{R}\right] $$
$$\phi=\tan^{-1}\left[\dfrac{\sqrt{\dfrac{4L}{ C}} -\dfrac{1}{\sqrt{\dfrac{4C}{L}}}}{R}\right] $$
Plug the known;
$$\phi=\tan^{-1}\left[\dfrac{\sqrt{\dfrac{4(0.01)}{ (1\times 10^{-9})}} -\dfrac{1}{\sqrt{\dfrac{ 4(1\times 10^{-9})}{ (0.01)}}}}{(100)}\right]=\bf 88.8^\circ $$
Plug into (2),
$$P_{\rm source}=\dfrac{\varepsilon_{\rm rms}^2\cos(88.8^\circ)}{\sqrt{\left(\sqrt{\dfrac{L}{4 C}} -\dfrac{1}{\sqrt{\dfrac{C}{4L}}}\right)^2+R^2}} $$
Plug the known;
$$P_{\rm source}=\dfrac{(10)^2 }{\sqrt{\left(\sqrt{\dfrac{4(0.01)}{ (1\times 10^{-9})}} -\dfrac{1}{\sqrt{\dfrac{ 4(1\times 10^{-9})}{ (0.01)}}}\right)^2+100^2}} $$
$$P_{\rm source}=\color{red}{\bf 4.44\times 10^{-4}}\;\rm W$$
