A ratio of the 5$$^{th}$$ term from the beginning to the 5$$^{th}$$ term from the end in the binomial expansion of $$\left(2^{1/3} + \frac{1}{2(3)^{1/3}}\right)^{10}$$ is

We have the binomial $$\left(2^{1/3}+\dfrac{1}{2\,(3)^{1/3}}\right)^{10}$$. Denote

$$a = 2^{1/3}, \qquad b = \dfrac{1}{2\,(3)^{1/3}}.$$

The general term in the expansion of $$(a+b)^{n}$$ is, by the Binomial Theorem, stated first as

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

where $$r=0,1,2,\dots ,n$$. Here $$n=10$$.

Because counting starts with $$r=0$$ for the first term, the 5th term from the beginning corresponds to $$r=4$$. Substituting $$n=10$$ and $$r=4$$ gives

$$T_{5\text{ (beg.)}} =^{10}C_4\,a^{\,10-4}\,b^{\,4}=^{10}C_4\,a^{6}\,b^{4}.$$

For the 5th term from the end we note that there are $$n+1=11$$ terms in total, so the 5th from the end is the $$11-5+1=7$$-th term from the beginning; thus it has index $$r=6$$. Substituting $$r=6$$ yields

$$T_{5\text{ (end)}} =^{10}C_6\,a^{\,10-6}\,b^{\,6}=^{10}C_6\,a^{4}\,b^{6}.$$

Now we form the required ratio:

$$\dfrac{T_{5\text{ (beg.)}}}{T_{5\text{ (end)}}}= \dfrac{^{10}C_4\,a^{6}\,b^{4}}{^{10}C_6\,a^{4}\,b^{6}}.$$

Because $$^{10}C_4 = ^{10}C_6 = 210,$$ the binomial coefficients cancel, giving

$$\dfrac{T_{5\text{ (beg.)}}}{T_{5\text{ (end)}}}= \dfrac{a^{6}\,b^{4}}{a^{4}\,b^{6}} = \dfrac{a^{2}}{b^{2}} = \left(\dfrac{a}{b}\right)^{2}.$$

So we must first evaluate $$\dfrac{a}{b}$$. Substituting $$a=2^{1/3}$$ and $$b=\dfrac{1}{2\,(3)^{1/3}}$$, we get

$$\dfrac{a}{b}=2^{1/3}\;\div\;\dfrac{1}{2\,(3)^{1/3}} = 2^{1/3}\times 2\,(3)^{1/3}=2\cdot 2^{1/3}\,(3)^{1/3}.$$

Observing that $$2^{1/3}(3)^{1/3}=(2\cdot 3)^{1/3}=(6)^{1/3},$$ we simplify further:

$$\dfrac{a}{b}= 2\,(6)^{1/3}.$$

Therefore

$$\left(\dfrac{a}{b}\right)^{2} = \left(2\,(6)^{1/3}\right)^{2}=4\,(6)^{2/3}.$$

Because $$(6)^{2/3}=\left(6^{2}\right)^{1/3}=36^{1/3},$$ we may write

$$\left(\dfrac{a}{b}\right)^{2}=4\,(36)^{1/3}.$$

Thus the ratio of the 5th term from the beginning to the 5th term from the end is

$$4\,(36)^{1/3}:1.$$

Hence, the correct answer is Option B.