The tube of a neon sign holds neon at a pressure of 1.3 × 10³ Pa and a temperature of 300 K…

Step 1. Number density. $n = \dfrac{p}{k_BT} = \dfrac{1.3 \times 10^3\ \text{Pa}}{1.38 \times 10^{-23}\ \text{J/K} \times 300\ \text{K}} = 3.14 \times 10^{23}\ \text{m}^{-3}$.

Step 2. Mean free path. The bottom line is $\sqrt{2}\,\pi \times (2.6 \times 10^{-10}\ \text{m})^2 \times 3.14 \times 10^{23}\ \text{m}^{-3} = 9.43 \times 10^{4}\ \text{m}^{-1}$. So $\lambda = \dfrac{1}{9.43 \times 10^{4}\ \text{m}^{-1}} = 1.06 \times 10^{-5}\ \text{m}$, which is $10.6$ micrometres.