Solution

This effect can be explained by the fact that as the pressure increases, the mean free path of the molecules, determined by their effective cross-section and concentration, decreases, and when it becomes less than the distance between the plates, the heat flux begins to be determined by this length and ceases to depend on the pressure, since $\lambda\sim\frac{1}{n}\Rightarrow\lambda\sim\frac{1}{p}$ , and the heat flux in the general case is $W\sim n\Rightarrow W\sim p$. We will use the formula from the solution to problem 5.4.16.
$$W_{1}=\frac{3}{2}kn\Delta T\sqrt{\frac{3kT}{m}}S$$

$$W_{2}=\frac{3}{2}\lambda kn\frac{\Delta T}{\delta}\sqrt{\frac{3kT}{m}}S$$

$$\frac{W_{1}}{W_{2}}=\frac{\delta}{\lambda}$$

$$\frac{W_{1}}{W_{2}}=4\delta \pi r^{2}n\Rightarrow$$
$$r = \frac{1}{2} (\frac{W_{1}}{W_{2}\pi n\delta})^\frac{1}{2}$$

However, for the evaluation, the coefficient $\frac{1}{2}$ must be discarded. Hence the final answer:

$$r \approx(\frac{W_{1}}{W_{2}\pi n\delta})^\frac{1}{2}$$