If the fourth term in the Binomial expansion

$\because \mathrm{T}_{4}=20 \times 8^{7}$

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

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

$\Rightarrow \frac{x^{\log _{8} x}}{x}=64$

Now, take $\log _{8}$ on both sides, we get

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

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

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