If the fourth term in the Binomial expansion of $$\left(\frac{2}{x} + x^{\log_8 x}\right)^6$$, $$(x > 0)$$ is $$20 \times 8^7$$, then a value of $$x$$ is:

The expansion of $$(a+b)^n$$ has its general, that is, $$(r+1)^{\text{th}}$$, term given by the binomial-theorem formula

$$T_{r+1}=\,^nC_r\,a^{\,n-r}\,b^{\,r}\,.$$

In the present problem we have

$$n=6,\qquad a=\dfrac{2}{x},\qquad b=x^{\log_8 x}.$$

The fourth term corresponds to $$r+1=4\; \Longrightarrow\; r=3$$, so according to the formula we write

$$T_4={}^{6}C_{3}\left(\dfrac{2}{x}\right)^{6-3}\left(x^{\log_8 x}\right)^{3}.$$

Calculating the binomial coefficient first,

$$^{6}C_{3}=\dfrac{6!}{3!\,3!}=20.$$

Now evaluate each power separately.

$$\left(\dfrac{2}{x}\right)^{6-3}=\left(\dfrac{2}{x}\right)^{3}=\dfrac{2^{3}}{x^{3}}=\dfrac{8}{x^{3}}.$$

$$\left(x^{\log_8 x}\right)^{3}=x^{3\log_8 x}\; \bigl(\text{because }(x^{m})^{k}=x^{mk}\bigr).$$

Multiplying all the factors we obtain

$$T_4=20\;\times\;\dfrac{8}{x^{3}}\;\times\;x^{3\log_8 x}=160\,x^{\,3\log_8 x-3}.$$

The statement of the question tells us that this fourth term is equal to $$20\times 8^{7}$$, so we set

$$160\,x^{\,3\log_8 x-3}=20\,8^{7}.$$

Dividing both sides by $$20$$ gives

$$8\,x^{\,3\log_8 x-3}=8^{7}.$$

Now divide both sides by $$8$$:

$$x^{\,3\log_8 x-3}=8^{6}.$$

The base $$8$$ is convenient, so let us represent $$x$$ itself as a power of $$8$$. Put

$$x=8^{k}\qquad(k\in\mathbb{R}).$$

Then

$$\log_8 x=\log_8\!\bigl(8^{k}\bigr)=k.$$

Substituting these into the exponent of $$x$$ in the previous equation, we get

$$x^{\,3\log_8 x-3}=x^{\,3k-3}=\bigl(8^{k}\bigr)^{3k-3}=8^{\,k\,(3k-3)}.$$

Therefore the equality becomes

$$8^{\,k\,(3k-3)}=8^{6}.$$

Since the bases are identical and positive (both equal to $$8$$), the exponents must be equal:

$$k\,(3k-3)=6.$$

Simplify this quadratic equation:

$$3k^{2}-3k-6=0 \;\Longrightarrow\; k^{2}-k-2=0\quad(\text{dividing by }3).$$

Factorising,

$$(k-2)(k+1)=0\;\Longrightarrow\;k=2\quad\text{or}\quad k=-1.$$

Each possible value of $$k$$ gives a corresponding value of $$x$$ via $$x=8^{k}$$:

$$k=2\;\Rightarrow\;x=8^{2}=64,$$

$$k=-1\;\Rightarrow\;x=8^{-1}=\dfrac{1}{8}.$$

Among the answer choices provided, only $$x=8^{2}$$ appears. Hence, the correct answer is Option D.