Answer
$8.67\times 10^9Hz\lt f\lt \infty$
Work Step by Step
We know that
$I_{rms}=\frac{V_{rms}}{2\pi fL}$
In the given scenario, we require $I_{rms}\lt 1mA$
$\implies \frac{V_{rms}}{2\pi fL}\lt 1mA$
We plug in the known values to obtain:
$\frac{12}{2\pi f\times 0.22\times 10^{-6}}\lt 1\times 10^{-3}$
This simplifies to:
$f\gt 8.67\times 10^9Hz$
Thus the required range is $8.67\times 10^9Hz\lt f\lt \infty$
