A vacuum chamber is 0.50 m across. Its walls are at 295 K…

Step 1. The pressure. We want $p$, so $p = \dfrac{k_BT}{\sqrt{2}\,\pi d^2\lambda}$:

The top line is $k_BT = 1.38 \times 10^{-23}\ \text{J/K} \times 295\ \text{K} = 4.07 \times 10^{-21}\ \text{J}$. The bottom line is $\sqrt{2}\,\pi \times (3.7 \times 10^{-10}\ \text{m})^2 \times 0.50\ \text{m} = 3.04 \times 10^{-19}\ \text{m}^3$. So

Step 2. Molecules left. $n = \dfrac{p}{k_BT} = \dfrac{0.0134\ \text{Pa}}{4.07 \times 10^{-21}\ \text{J}} = 3.29 \times 10^{18}\ \text{m}^{-3}$. In each cubic centimetre that is $3.29 \times 10^{12}$ molecules.