Answer
$\approx 84.9$
Work Step by Step
We know that the mean free path is given by
$$\lambda=\dfrac{1}{4\sqrt{2}\pi r(\frac{N}{V})}$$
And are given that the mean free path is ten times the atomic diameter.
Hence,
$$\lambda=10(2r)=20r$$
Plugging into (1) and then solving for $\frac{V}{N}$
$$20r=\dfrac{V}{N}\dfrac{1}{4\sqrt{2}\pi r }$$
Hence,
$$\dfrac{V}{N}=20r(4\sqrt{2}\pi r)$$
$$\dfrac{V}{N}=80\sqrt{2}\pi r^3$$
We need to find the ratio of the volume per atom $(V/N)$ to the volume of an atom.
$$\dfrac{V/N}{V_{atom}}=\dfrac{80\sqrt{2} \color{red}{\bf\not} \pi \color{red}{\bf\not} r^3}{\frac{4}{3} \color{red}{\bf\not} \pi \color{red}{\bf\not} r^3}=60\sqrt2$$
$$\dfrac{V/N}{V_{atom}}=\color{red}{\bf 84.85}$$
