Diffusion of  ${\text{CH}}_4\left(g\right)$ and  ${\text{O}}_2\left(g\right)$ occurs under similar conditions, then the ratio of  their rates of diffusion is 

From Graham’s law of diffusion : 
 $\frac{r_{C{H_4}_4}}{r_{O_2}}=\sqrt{\frac{M_{O_2}}{M_{C{H}_4}}}$
Where $r_{C{H}_4}=$ rate of diffusion of $C{H}_4$ gas 
$r_{O_2=}$  rate of diffusion of $O_2$ gas
$M_{C{H}_4}=$ molecular mass of  $C{H}_4$
$M_{O_2}=$  molecular mass of $O_2$ gas 

$\therefore \frac{r_{C{H}_4}}{r_{O_2}}=\sqrt{\frac{32}{16}}=\sqrt{2}=1.414$