Answer
See the detailed answer below.
Work Step by Step
a) From the given definition of $Q$,
$$Q=\dfrac{V_L}{V_R}=\dfrac{IX_L}{IR} $$
$$Q= \dfrac{ X_L}{ R} =\dfrac{2\pi f L}{R}$$
And we know that the resonance frequency is given by
$$f=\dfrac{1}{2\pi}\sqrt{\dfrac{1}{LC}}$$
So that;
$$Q= \dfrac{2\pi L}{R}\dfrac{1}{2\pi}\sqrt{\dfrac{1}{LC}}$$
$$Q= \dfrac{ L}{R} \sqrt{\dfrac{1}{LC}}= \dfrac{ 1}{R}\sqrt{\dfrac{L^2}{LC}}$$
$$\boxed{Q= \dfrac{ 1}{R}\sqrt{\dfrac{L }{ C}}}$$
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b)
First, we need to find $L$;
$$f=\dfrac{1}{2\pi}\sqrt{\dfrac{1}{LC}}$$
$$(2\pi f)^2= \dfrac{1}{LC}$$
$$L = \dfrac{1}{(2\pi f)^2C}$$
Plugging the given;
$$L = \dfrac{1}{(2\pi \cdot 1.0\times10^6)^2\cdot 0.01\times10^{-6}}$$
$$L =\color{red}{\bf 2.53\times10^{-6}}\;\rm H$$
Now we need to find $R$ by solving the boxed formula above for it.
$$R= \dfrac{ 1}{Q}\sqrt{\dfrac{L }{ C}}$$
Plugging the given;
$$R= \dfrac{ 1}{650}\sqrt{\dfrac{ 2.53\times10^{-6} }{ 0.01\times10^{-6}}}$$
$$R= \color{red}{\bf0.0245}\;\rm \Omega$$