Treat a nitrogen molecule as a ball of diameter 0.30 nm…

Step 1. One molecule. $\tfrac{\pi}{6}d^3 = \tfrac{\pi}{6} \times (0.30 \times 10^{-9}\ \text{m})^3 = 1.41 \times 10^{-29}\ \text{m}^3$.

Step 2. At $101.3$ kPa. $\dfrac{k_BT}{p} = 4.09 \times 10^{-26}\ \text{m}^3$. The fraction is $\dfrac{1.41 \times 10^{-29}\ \text{m}^3}{4.09 \times 10^{-26}\ \text{m}^3} = 3.5 \times 10^{-4}$, that is $0.035\ \%$.

Step 3. At $20.0$ MPa. The pressure is $197$ times larger, so the room for each molecule is $197$ times smaller: $2.07 \times 10^{-28}\ \text{m}^3$. The fraction is $0.068$, that is $6.8\ \%$.