The solution of this question is given by assuming step $\left(\mathrm{i}\right)$ to be reversible which is not given in question

$\mathrm{Overall} \mathrm{rate} = \mathrm{Rate} \mathrm{of} \mathrm{slowest} \mathrm{step} \left(\mathrm{ii}\right)$

$=\mathrm{k}\left[\mathrm{X}\right]\mathrm{ }\left[{\mathrm{Y}}_2\right]$ ......$\left(\mathrm{i}\right)$

K$=$rate constant of step $\left(\mathrm{ii}\right)$

Assuming step $\left(\mathrm{i}\right)$ to be reversible, its equilibrium constant,

$k_{eq}=\frac{{\left[X\right]}^2}{\left[X_2\right]} \Rightarrow \left[X\right]=k_{eq}^{\frac{1}{2}} {\left[X_2\right]}^{\frac{1}{2}}$ .....$\left(\mathrm{ii}\right)$

Put $\left(\mathrm{ii}\right)$ in $\left(\mathrm{i}\right)$

Rate $=k{k}_{eq}^{\frac{1}{2}} {\left[X_2\right]}^{\frac{1}{2}} \left[Y_2\right]$

Overall order $=\frac{1}{2}+1=\frac{3}{2}$