A circuit element X when connected to an AC supply of peak voltage...

As current is in phase with the applied voltage, element $X$ should be resistive with $R = \dfrac{{V}_{0}}{{I}_{0}} = \dfrac{100}{5} = 20Ω$.

As current lags behind voltage by $90 °$, element $Y$ should be inductive with ${X}_{L} = \dfrac{{V}_{0}}{{I}_{0}} = \dfrac{100}{5} = 20Ω$

When $X$ and $Y$ are connector in series,

Impedance, $Z = \sqrt{{X}_{L}^{2} + {R}^{2}} = \sqrt{{20}^{2} + {20}^{2}} = 20\sqrt{2}Ω$

Now, peak current, ${I}_{0} = \dfrac{{V}_{0}}{Z} = \dfrac{100}{20\sqrt{2}} = \dfrac{5}{\sqrt{2}}A$

Thus, the rms value of current is ${I}_{rms} = \dfrac{{I}_{0}}{\sqrt{2}} = \dfrac{5}{2}A$