The general term in the binomial expansion ${\left(a+b\right)}^n$ is $T_{r+1}={}^{n}C_r{a}^{n-r}{b}^r.$

Given, the fourth term in the expansion of ${\left(\frac{2}{x}+x^{{\log }_{8}x}\right)}^6$ is 

$T_4=20\times 8^7$

$\Rightarrow {}^{6}C_3{\left(\frac{2}{x}\right)}^3{\left(x^{{\log }_{8}x}\right)}^3=20\times 8^7$

$\Rightarrow \frac{6!}{3!·3!}{\left(\frac{2}{x}\right)}^3{\left(x^{{\log }_{8}x}\right)}^3=20\times {\left(2^3\right)}^7$

$\Rightarrow \left(\frac{6\times 5\times 4\times 3!}{3\times 2\times 1·3!}\right){\left(\frac{2}{x}\right)}^3{\left(x^{{\log }_{8}x}\right)}^3=20\times 2^{21}$

$\Rightarrow \left(20\right){\left(\frac{2}{x}\right)}^3{\left(x^{{\log }_{8}x}\right)}^3=20\times 2^{21}$

$\Rightarrow {\left[\left(\frac{2}{x}\right)\left(x^{{\log }_{8}x}\right)\right]}^3={\left(2^7\right)}^3$

$\Rightarrow \left(\frac{2}{x}\right)\left(x^{{\log }_{8}x}\right)=2^7$

$\Rightarrow \left(\frac{1}{x}\right)\left(x^{{\log }_{8}x}\right)=2^6$

$\Rightarrow x^{{\log }_{8}x-1}=8^2$

Taking $\log$ both side to the base $8$

$\Rightarrow {\log }_8\left(x^{{\log }_{8}x-1}\right)={\log }_8\left(8^2\right)$

Using ${\log }_a{m}^n=n{\log }_{a}m,$ we get

$\Rightarrow ({\log }_{8}x-1)\left({\log }_{8}x\right)=2$

$\Rightarrow \left(t-1\right)\left(t\right)=2$                (Let ${\log }_8\left(x\right)=t$)

$\Rightarrow t^2-t-2=0$

$\Rightarrow t=2$ or $-1$

$\Rightarrow {\log }_{8}x=2$ or $-1$

If ${\log }_{a}x=b, \Rightarrow x=a^b,$

$\Rightarrow x=8^2$ or $8^{-1}$

$\Rightarrow x=64$ or $\frac{1}{8}.$