At what temperature will the rate of diffusion of ${\mathrm{N}}_2$ be $1.6$ times the rate of diffusion of ${\mathrm{SO}}_2$ at $27°\mathrm{C}$ ?

We know that the rate of diffusion of a has is directly proportional to  $\sqrt{\frac{\mathrm{T}}{\mathrm{M}}}$.

Representing the given system mathematically, we have:

$\frac{{\mathrm{R}}_{{\mathrm{N}}_2}}{{\mathrm{R}}_{{\mathrm{SO}}_2}}=\sqrt{\frac{{\mathrm{T}}_{{\mathrm{N}}_2}}{{\mathrm{M}}_{{\mathrm{N}}_2}}\times \frac{{\mathrm{M}}_{{\mathrm{SO}}_2}}{{ \mathrm{T}}_{{\mathrm{SO}}_2}}}$

Substituting values:

$1.6=\sqrt{\frac{{\mathrm{T}}_{{\mathrm{N}}_2}}{28}\times \frac{64}{300}}$

$\frac{{\mathrm{T}}_{{\mathrm{N}}_2}\times 64}{8400}=2.56$

${\mathrm{T}}_{{\mathrm{N}}_2}=336 \mathrm{K}=63°\mathrm{C}$

Hence, the required answer is $63°\mathrm{C}$.