Explanation
Hence, at resonance:
$$X_L = X_C$$
So the voltage across rthe unknown inductor is $$200V$$ as same as the voltage across the capacitor.
Current is maximum at resonance
$$\displaystyle \Rightarrow \omega^2 = \frac{1}{LC} \Rightarrow L = \frac{1}{\omega^2 C}$$
$$\displaystyle = \frac{1}{(1000)^2 (10 \times 10^{-6})} = 0.1 H =100 mH$$
Please disable the adBlock and continue. Thank you.