Answer
a) ${\bf 11.6}\;\rm pF $
b) ${\bf 1.49}\;\rm m\Omega$
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know, for $RLC$-circuits, that the resonance frequency occurs when
$$X_L=X_C$$
$$2\pi f L=\dfrac{1}{2\pi f C}$$
Hence,
$$C=\dfrac{1}{4\pi f^2 L}$$
Plug the known;
$$C=\dfrac{1}{4\pi (104.3\times 10^6)^2 (0.20\times 10^{-6})}$$
$$C=\color{red}{\bf 1.16\times 10^{-11}}\;\rm F$$
$$\color{blue}{\bf [b]}$$
The peak current of the original station is given by
$$I=\dfrac{\varepsilon_0}{Z} =\dfrac{\varepsilon_0}{\sqrt{(X_L-X_C)^2+R^2}}$$
where at the resonance frequency $X_L=X_C$, so $Z=R$
$$I =\dfrac{\varepsilon_0}{R}$$
The current of the out-of-tune station is
$$I'=\dfrac{0.1}{100}I=\dfrac{0.1}{100}\dfrac{\varepsilon_0}{R}$$
where $I'=\varepsilon_0/Z=\varepsilon_0/\sqrt{(X_L-X_C)^2+R^2}$
$$ \dfrac{ \color{red}{\bf\not} \varepsilon_0}{\sqrt{(X_L-X_C)^2+R^2}} =\dfrac{0.1}{100}\dfrac{ \color{red}{\bf\not} \varepsilon_0}{R}$$
Squaring both sides;
$$ \dfrac{ 1}{ (X_L-X_C)^2+R^2} = \dfrac{ 10^{-6}}{R^2}$$
Hence,
$$R^2=10^{-6}\left[(X_L-X_C)^2+R^2\right]$$
$$10^{ 6}R^2-R^2= (X_L-X_C)^2 $$
$$R =\sqrt{ \dfrac{(X_L-X_C)^2 }{(10^{ 6} -1)}}$$
$$R =\dfrac{|X_L-X_C|}{\sqrt{ (10^{ 6} -1)}}$$
where $X_L=\omega L=2\pi fL$, and $X_C=1/2\pi fC$
$$R =\dfrac{\left|2\pi f'L-\dfrac{1}{2\pi f' C}\right|}{\sqrt{ (10^{ 6} -1)}}$$
Plug the known;
$$R =\dfrac{\left|2\pi (103.9\times 10^{6})(0.2\times 10^{-6})-\dfrac{1}{2\pi (103.9\times 10^{6}) (1.16\times 10^{-11})}\right|}{\sqrt{ (10^{ 6} -1)}}$$
$$R=\color{red}{\bf 1.49}\;\rm m\Omega$$