The percentage by mole of ${\mathrm{NO}}_2$ in a mixture of ${\mathrm{NO}}_2 \left(\mathrm{g}\right)$and  $\mathrm{NO}\left(\mathrm{g}\right)$having average molecular mass $34$ is:

Let x represent the percentage by mole of ${\mathrm{NO}}_2$ in the mixture.
The molecular masses of ${\mathrm{NO}}_2$ and $\mathrm{NO}$ are 46 g/mol and 30 g/mol respectively.

$\begin{array}{l}{M}_{avg}=\frac{x\cdot M{\mathrm{NO}}_2+(100-x)M_{\mathrm{NO}}}{100}\\ 34=\frac{x\times 46+(100-x)30}{100}\\ x=25\mathrm{\%}\end{array}$

Thus, the average molecular mass $34$ is 25%.