If the coefficients of $x^{-2}$  and  $x^{-4}$   in the expansion of  ${(x^{\frac{1}{3}}+\frac{1}{2x^{\frac{1}{3}}})}^{18}$ (x>0) are m and n respectively then $\frac{m}{n}$  is divisible by

$T_{r+1}={}^{18}C_r{\left(x^{\frac{1}{3}}\right)}^{18-r}\frac{1}{{\left(2x^{\frac{1}{3}}\right)}^r}={}^{18}C_r\frac{x^{6-\frac{2r}{3}}}{2^r}$

For Coefficient of  $x^{-2}$ is  $6-\frac{2r}{3}=-2\Rightarrow r=12$
Coefficient $x^{-4}$  is  $6-\frac{2r}{3}=-4\Rightarrow r=15$

$\therefore \frac{m}{n}=\frac{{}^{18}C_{12}\frac{1}{2^{12}}}{{}^{18}C_{15}\frac{1}{2^{15}}}=182$