If the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of $$\sqrt[4]{2} + \dfrac{1}{\sqrt[4]{3}}  ^n$$ is $$\sqrt{6}:1$$, then the third term from the beginning is:

Let $$a = \sqrt[4]{2} = 2^{1/4}$$ and $$b = \frac{1}{\sqrt[4]{3}} = 3^{-1/4}$$.

The fifth term from the beginning is: $$T_5 = \binom{n}{4} a^{n-4} b^4$$

The fifth term from the end is: $$T_{n-3} = \binom{n}{4} a^4 b^{n-4}$$

The ratio is:

$$ \frac{T_5}{T_{n-3}} = \frac{a^{n-4} b^4}{a^4 b^{n-4}} = \left(\frac{a}{b}\right)^{n-8} = \left(2^{1/4} \cdot 3^{1/4}\right)^{n-8} = 6^{(n-8)/4} $$

Given this ratio equals $$\sqrt{6} = 6^{1/2}$$:

$$ \frac{n-8}{4} = \frac{1}{2} \implies n - 8 = 2 \implies n = 10 $$

The third term from the beginning is:

$$ T_3 = \binom{10}{2} \cdot (2^{1/4})^8 \cdot (3^{-1/4})^2 = 45 \cdot 2^2 \cdot 3^{-1/2} = 45 \cdot \frac{4}{\sqrt{3}} = \frac{180}{\sqrt{3}} = 60\sqrt{3} $$

The correct answer is $$60\sqrt{3}$$.