Given, $x=0.20 \mathrm{m},\mathrm{ }y=0.20 \mathrm{m},\mathrm{ }u=\mathrm{1.8 }{\mathrm{m }\mathrm{s}}^{-1}$

Let the ball strike the $n$th step of stairs,

Vertical distance travelled $=\text{n}\text{y}$

Using equation of motion in y-direction

$s=ut+\frac{1}{2}a{t}^2$

$⟹-ny=\mathrm{0-}\frac{1}{2}g{t}^2$        ...(1)

Horizontal distance travelled $=nx$

$⟹\text{n}\text{x}=ut$

$⟹t=\frac{nx}{u}$       ...(2)

using (1) and (2)

$ny=\frac{1}{2}g{\left(\frac{nx}{u}\right)}^2$

$\mathrm{o}\mathrm{r}\mathrm{ }n=\frac{2u^2}{g}\frac{y}{x^2}$ $=3.3⟹4^{th} stair$