Answer
Predicted mean free path = $6*10^{12}$
Work Step by Step
Mean free path = $\frac{1}{\sqrt 2*pi*d^{2}*N/V}$
Here, $d$ id the diameter, $N$ is the number of molecule and $V$ is the volume
Given that, $d=2.0*10^{-10}$ and $N/V=1*10^{6}molecules/m^{3}$
So, Mean free path= $\frac{1}{\sqrt 2*pi*(2.0*10^{-10})*1*10^{-6}}=6*10^{12}m$
