Solution

In the classical case, the heat flux is determined by the formula:

$$q=-\frac{1}{3}\lambda\left\langle \upsilon \right\rangle \frac{i}{2}kn\frac{\partial T}{\partial x}S$$

However, for the sake of clarity, it would be fair to ignore the coefficient $\frac{1}{3}$. It should also be noted that for $\lambda\gg \delta$, the mean free path of the molecules is determined by the distance $\delta$. Also, given the smallness of $\delta$, the derivative can be considered equal to $-\frac{\Delta T}{\delta}$. The gas is monatomic, meaning $i=3$. We'll take the average velocity to be equal to the root-mean-square value: $\left\langle \upsilon \right\rangle = \sqrt{\frac{3kT}{\mu}}$. Hence, the final formula for the flux density is:

$$\omega=\frac{q}{S}=\frac{3}{2}kn\Delta T\sqrt{\frac{3kT}{\mu}}$$