Use the half-power condition for a series LCR circuit

In a series LCR circuit, the current amplitude is $I = \frac{V}{\sqrt{R^2 + \left(\omega L - \frac{1}{\omega C}\right)^2}}.$

The current is maximum at resonance, where $\omega_0 L = \frac{1}{\omega_0 C},$ so $I_{\max} = \frac{V}{R}.$

We are given the two angular frequencies at which $I = \frac{I_{\max}}{\sqrt{2}}.$

These are the half-power frequencies $\omega_1$ and $\omega_2$.

Bandwidth relation

For a series LCR circuit, the half-power angular frequencies satisfy $\omega_2 - \omega_1 = \frac{R}{L}.$

Given: $\omega_1 = 212\ \text{rad s}^{-1}, \qquad \omega_2 = 232\ \text{rad s}^{-1}.$

So, $\omega_2 - \omega_1 = 232 - 212 = 20\ \text{rad s}^{-1}.$

Hence, $\frac{R}{L} = 20.$

Substitute the resistance

Given $R = 5\ \Omega.$

Therefore, $\frac{5}{L} = 20.$

So, $L = \frac{5}{20} = 0.25\ \text{H}.$

Convert into mH

$0.25\ \text{H} = 250\ \text{mH}.$

Final answer

$\boxed{250\ \text{mH}}$