A circuit element X when connected to an AC supply of peak voltage 100   V gives a peak current of 5…

As current is in phase with the applied voltage, element $X$ should be resistive with $R=\frac{V_0}{I_0}=\frac{100}{5}=20 \mathrm{\Omega }$.

As current lags behind voltage by $90°$, element $Y$ should be inductive with $X_L=\frac{V_0}{I_0}=\frac{100}{5}=20 \mathrm{\Omega }$

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

Impedance, $Z=\sqrt{X_L^2+R^2}=\sqrt{{20}^2+{20}^2}=20\sqrt{2} \mathrm{\Omega }$

Now, peak current, $I_0=\frac{V_0}{Z}=\frac{100}{20\sqrt{2}}=\frac{5}{\sqrt{2}} \mathrm{A}$

Thus, the rms value of current is $I_{\mathrm{rms}}=\frac{I_0}{\sqrt{2}}=\frac{5}{2} \mathrm{A}$