Nitrogen has a critical temperature of 126.2 K and a critical pressure of 33.9 bar…

Step 1. The constant $b$. $b = \dfrac{RT_c}{8P_c} = \dfrac{0.08314 \times 126.2}{8 \times 33.9} = 0.0387\ \text{L/mol}$.

Step 2. The constant $a$. $a = \dfrac{27R^2T_c^2}{64P_c} = \dfrac{27 \times 0.08314^2 \times 126.2^2}{64 \times 33.9} = 1.370\ \text{L}^2\ \text{bar mol}^{-2}$.

Step 3. The radius. We want $r$ from $b = 4N_A\dfrac{4}{3}\pi r^3 = \dfrac{16}{3}\pi N_Ar^3$. So $r^3 = \dfrac{3b}{16\pi N_A} = \dfrac{3 \times 0.0387 \times 10^{-3}\ \text{m}^3}{16\pi \times 6.022 \times 10^{23}} = 3.84 \times 10^{-30}\ \text{m}^3$, and $r = 1.57 \times 10^{-10}\ \text{m} = 0.157\ \text{nm}$.

Step 4. The Boyle temperature. $T_B = \dfrac{a}{Rb} = \dfrac{1.370}{0.08314 \times 0.0387} = 426\ \text{K}$. It also equals $\dfrac{27}{8}T_c = 426\ \text{K}$.

Step 5. At $300\ \text{K}$. $300\ \text{K}$ is below $426\ \text{K}$, so at low pressure $Z < 1$ for nitrogen. Attraction wins at low pressure.