$Y=\frac{K T}{\sqrt{2} \pi \sigma^{2} p}$

-wherein

$\sigma=$ Diameter of the molecule

$p=$ pressure of the gas

$T=$ temperature

$K=$ Boltzmann's Constant

Let intermolecular distance be $D$ then in a volume $\frac{4 \pi}{3} D^{3}$ there is only one

$\frac{4 \pi}{3} D^{3} P=\frac{1}{N_{A}}=R_{T} \quad$ or $\quad D=\left(\frac{3 R T}{4 \pi N_{A} P}\right)^{1 / 3}$

$\begin{aligned} \text { Put } P=& 4 \times 10^{-10} P a, R=83, N_{A}=6 \times 10^{23} \\ T=& 300 \mathrm{K} \end{aligned}$

$D=0.2 \mathrm{mm}$