Answer
a) ${\bf 4.873\times 10^{-7}}\;\rm H$
b) ${\bf 10.9 }\;\rm \Omega$
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,
$$L=\dfrac{1}{4\pi f^2 C}$$
Plug the known; where $f=\dfrac{f_{max}+f_{min}}{2}=\dfrac{60+54}{2}=\bf 57\;\rm MHz$
$$L=\dfrac{1}{4\pi^2 (57\times 10^6)^2 (16\times 10^{-12})}$$
$$L=\color{red}{\bf 4.873\times 10^{-7}}\;\rm H$$
$$\color{blue}{\bf [b]}$$
The peak current at the resonance frequency is
$$I=\dfrac{\varepsilon_0}{Z} =\dfrac{\varepsilon_0}{\sqrt{(X_L-X_C)^2+R^2}}$$
where at the resonance frequency $X_L=X_C$, so $Z=R$
$$I =\dfrac{\varepsilon_0}{R}$$
The current at the end frequency is
$$I'=\dfrac{1}{2}I=\dfrac{1}{2}\dfrac{\varepsilon_0}{R}$$
where $I'=\varepsilon_0/Z=\varepsilon_0/\sqrt{(X_L-X_C)^2+R^2}$
$$ \dfrac{ \color{red}{\bf\not} \varepsilon_0}{\sqrt{(X_L-X_C)^2+R^2}} =\dfrac{1}{2}\dfrac{ \color{red}{\bf\not} \varepsilon_0}{R}$$
Squaring both sides;
$$ \dfrac{ 1}{ (X_L-X_C)^2+R^2} = \dfrac{ 1}{4R^2}$$
Hence,
$$4R^2= (X_L-X_C)^2+R^2 $$
$$3R^2= (X_L-X_C)^2 $$
$$R =\sqrt{ \dfrac{(X_L-X_C)^2 }{3}}$$
$$R =\dfrac{|X_L-X_C|}{\sqrt{ 3}}$$
where $X_L=\omega L=2\pi fL$, and $X_C=1/2\pi fC$
$$R =\dfrac{\left|2\pi f L-\dfrac{1}{2\pi f C}\right|}{\sqrt{3}}$$
Plug the known for $f_{\rm max}$;
$$R =\dfrac{\left|2\pi (60\times 10^6)(4.873\times 10^{-7})-\dfrac{1}{2\pi (60\times 10^6)(16\times 10^{-12})}\right|}{\sqrt{3}}$$
$$R_1=\bf 10.35\;\rm \Omega$$
Plug the known for $f_{\rm min}$;
$$R_2 =\dfrac{\left|2\pi (54\times 10^6)(4.873\times 10^{-7})-\dfrac{1}{2\pi (54\times 10^6)(16\times 10^{-12})}\right|}{\sqrt{3}}$$
$$R_2=\bf 10.9\;\rm \Omega$$
It is obvious now that the minimum possible value of the circuit’s resistance is 10.9 Ohm.
$$R_{\rm min}=\color{red}{\bf 10.9 }\;\rm \Omega$$
