A tuner can use one of two coils. Both have the same inductance…

Step 1. The quality factors. $Q = \dfrac{\omega_0L}{R}$ and $\omega_0$ and $L$ are the same. So $Q \propto \dfrac{1}{R}$ and $\dfrac{Q_1}{Q_2} = \dfrac{24\ \Omega}{6\ \Omega} = 4$.

Step 2. The largest currents. At resonance $I_{max} = \dfrac{V}{R}$. So $\dfrac{I_1}{I_2} = \dfrac{R_2}{R_1} = \dfrac{24\ \Omega}{6\ \Omega} = 4$.

Step 3. The bandwidths. $\Delta f = \dfrac{f_0}{Q}$ and $f_0$ is the same. So $\dfrac{\Delta f_1}{\Delta f_2} = \dfrac{Q_2}{Q_1} = \dfrac{1}{4} = 0.25$.