Step 1: Find the mass of nitrogen molecules.
The temperature of the nitrogen molecule, T = 300 K
The atomic mass of nitrogen = 14.0076 u
Hence, mass of the nitrogen molecule, m = 2 x 14.0076 = 28.0152 u
But 1 u= 1.66 x ${10}^{-27}$kg
So, m= 28.0152 x1.66 x ${10}^{-27}$ kg
Step 2: Find root mean square speed.
Planck's constant, h = 6.63 x ${10}^{-34}$ Js
Boltzmann constant, k = 1.38 x ${10}^{-23} \mathrm{J} {\mathrm{K}}^{-1}$
We have the expression that relates the mean kinetic energy kT)of the nitrogen molecule with the root mean square speed as:

$\begin{array}{l}\frac{1}{2}m{v}_{\mathrm{rms}\mathrm{}}^2=\frac{3}{2}kT\\ v_{r\mathrm{ms}}=\sqrt{\frac{3kT}{m}}\end{array}$

Step 3:
Hence, the de Broglie wavelength of the nitrogen molecule is given as:
$\begin{array}{l}\lambda =\frac{h}{m{v}_{rms}}=\frac{h}{\sqrt{3mkT}}\\ =\frac{6.63\times {10}^{-34}}{\sqrt{3\times 28.0152\times 1.66\times {10}^{-27}\times 1.38\times {10}^{-23}\times 300}}\\ =0.028\times {10}^{-9}m\\ =0.028nm\end{array}$
Therefore, the de Broglie wavelength of the nitrogen molecule is 0.028 nm.