Since the voltage across the inductor and capacitor is same, the circuit is in resonance. Therefore, the current in the circuit will be,

$i=\frac{V}{R}=\frac{220}{5}=44 \mathrm{A}$.

The voltage across the inductor is double that across the resistor. Therefore, $V_L=2V_R=440 \mathrm{V}$.

We know that $V_L=i{X}_L=i\times 2\pi fL\Rightarrow L=\frac{440}{44\times 100\pi }=\frac{1}{10\pi } \mathrm{H}=\frac{100}{\mathrm{\pi }}\mathrm{mH}$

Therefore, $K=\frac{1}{100}=0.01\approx 0$.