How many atoms of helium are in the chamber?
You already mentioned the equation for the rate of collisions of gas molecules with a section of wall:

$$Z_\mathrm w=\frac14\frac NV\sqrt{\frac{8RT}{\pi M}}A$$

We know the values of
rate $Z_\mathrm w=2.25\times10^{15}\ \mathrm{s^{-1}}$,
volume $V=22.7\ \mathrm l=0.0227\ \mathrm{m^3}$,
molar gas constant $R=8.314462618\ \mathrm{J\ mol^{-1}\ K^{-1}}$,
temperature $T=298\ \mathrm K$,
molar mass of helium $M=4.003\ \mathrm{g\ mol^{-1}}=0.004003\ \mathrm{kg\ mol^{-1}}$, and
area $A=1.63\times10^{-5}\ \mathrm{mm^2}=1.63\times10^{-11}\ \mathrm{m^2}$.

Thus, the number $N$ of helium atoms is

$$\begin{align}
N&amp;=\frac{4Z_\mathrm wV}{\sqrt{\dfrac{8RT}{\pi M}}A}\\[6pt]
&amp;=\frac{4\times2.25\times10^{15}\ \mathrm{s^{-1}}\times 0.0227\ \mathrm{m^3}}{\sqrt{\dfrac{8\times8.314462618\ \mathrm{J\ mol^{-1}\ K^{-1}} \times 298\ \mathrm K}{\pi \times 0.004003\ \mathrm{kg\ mol^{-1}}}}\times1.63\times10^{-11}\ \mathrm{m^2}}\\[6pt]
&amp;=9.98\times10^{21}
\end{align}$$