For any ideal gas, density $\rho = \large\frac{PM}{RT}$

So, for densities to be equal under similar conditions of pressure, $\large\frac{M}{T}$ needs to be the same.

For $N_2O$ and $NO_2, \; \large\frac{44}{273} $$= \large\frac{46}{273+12}$

Hence this is the correct answer. (We can verify the values of $\large\frac{M}{T}$ for the other combinations and we'll see that they are not the same).