A radio has a coil of 160 µH. Its variable capacitor must tune stations from 540 kHz to 1600 kHz…

Step 1. At $540$ kHz. $\omega_0 = 2\pi \times 540 \times 10^3\ \text{Hz} = 3.39 \times 10^6\ \text{rad/s}$. We want $C$, so $C = \dfrac{1}{\omega_0^2L}$. The bottom is $(3.39 \times 10^6\ \text{s}^{-1})^2 \times 160 \times 10^{-6}\ \text{H} = 1.84 \times 10^{9}$ in SI units. So $C = 5.43 \times 10^{-10}\ \text{F} = 543\ \text{pF}$.

Step 2. At $1600$ kHz. $\omega_0 = 2\pi \times 1600 \times 10^3\ \text{Hz} = 1.01 \times 10^7\ \text{rad/s}$. The bottom is $(1.01 \times 10^7\ \text{s}^{-1})^2 \times 160 \times 10^{-6}\ \text{H} = 1.62 \times 10^{10}$ in SI units. So $C = 6.18 \times 10^{-11}\ \text{F} = 61.8\ \text{pF}$.

Step 3. The ratio. $\dfrac{542.9\ \text{pF}}{61.84\ \text{pF}} = 8.78$. This equals $\left(\dfrac{1600}{540}\right)^2$, because $C \propto 1/f_0^2$.