An inductor of 0 . 5mH , a capacitor of 20μ F and resistance of 20Ω...

Here, emf and current are in phase, so the circuit is either pure resistive circuit or case of resonance,

${X}_{L} = {X}_{C}$

Thus, the impedance of circuit is $Z = \sqrt{{R}^{2} + {\left({X}_{L} - {X}_{C}\right)}^{2}} = R = 20Ω$

Now, rms value of current is ${i}_{rms} = \dfrac{220}{20} = 11A$

Using the relation between peak and rms value of current, we have peak value ${i}_{0 =}{i}_{rms}\sqrt{2} = 11\sqrt{2} = \sqrt{242}A$.

Hence, the value of $x = 242$.