If the third term in the binomial expansion of $$(1 + x^{\log_2 x})^5$$ equals 2560, then a possible value of $$x$$ is:

We recall the binomial theorem, which says that in the expansion of $$(a+b)^n$$ the general term (often called the $$(r+1)^{\text{th}}$$ term, starting the count from $$r=0$$) is

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

In the given expression $$(1+x^{\log_{2}x})^{5}$$ we have $$a=1,\;b=x^{\log_{2}x},\;n=5.$$

The third term corresponds to $$r=2$$ (because for the first term $$r=0$$, for the second $$r=1$$, and for the third $$r=2$$). Substituting these values in the formula we obtain

$$T_{3}=\,\,{^{5}C_{2}}\,(1)^{5-2}\,\bigl(x^{\log_{2}x}\bigr)^{2}.$$

Since any power of 1 is 1, the factor $$(1)^{5-2}$$ is simply 1, and we are left with

$$T_{3}=\,\,{^{5}C_{2}}\;x^{2\log_{2}x}.$$

We know that $${^{5}C_{2}}=\dfrac{5!}{2!\,3!}=10,$$ so

$$T_{3}=10\,x^{2\log_{2}x}.$$

The problem states that this third term equals 2560, so we set

$$10\,x^{2\log_{2}x}=2560.$$

Dividing both sides by 10 gives

$$x^{2\log_{2}x}=256.$$

Next we express 256 as a power of 2. We know $$256=2^{8},$$ so the equation becomes

$$x^{2\log_{2}x}=2^{8}.$$

To simplify the left-hand side, let us write $$\log_{2}x=k.$$ Then $$x=2^{k}.$$ Substituting this into the left-hand side gives

$$x^{2\log_{2}x}=(2^{k})^{2k}=2^{k\cdot2k}=2^{2k^{2}}.$$

Therefore our equation is now

$$2^{2k^{2}}=2^{8}.$$

Because the bases are equal, we equate the exponents:

$$2k^{2}=8.$$

Dividing by 2 yields

$$k^{2}=4.$$

Taking square roots,

$$k=\pm2.$$

Recall that $$k=\log_{2}x,$$ so we have two cases:

1. $$\log_{2}x=2\;\;\Longrightarrow\;\;x=2^{2}=4.$$

2. $$\log_{2}x=-2\;\;\Longrightarrow\;\;x=2^{-2}=\dfrac{1}{4}.$$

Both values are positive and hence admissible, but only one of them appears among the given options. Inspecting the options, we see that $$\dfrac{1}{4}$$ is listed, whereas 4 is not.

Hence, the correct answer is Option 4.