1. Let $a=\sqrt[4]{2}=2^{1/4},\qquad b=\frac{1}{\sqrt[4]{3}}=3^{-1/4}.$ Then the expansion is $(a+b)^n.$

2. The general term from the beginning is $T_{r+1}=\binom{n}{r}a^{n-r}b^r.$

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

3. The fifth term from the end of a binomial expansion is the same as the term number $(n-4+1)=(n-3)\text{rd from the beginning},$ but more directly, it is $T_{n-3}=\binom{n}{4}a^4b^{n-4}.$

Thus, \frac{T_5}{\text{5th term from end}}=rac{\binom{n}{4}a^{n-4}b^4}{\binom{n}{4}a^4b^{n-4}}=a^{n-8}b^{8-n}=\left(\frac{a}{b}\right)^{n-8}.

4. Now $\frac{a}{b}=\frac{2^{1/4}}{3^{-1/4}}=(2\cdot 3)^{1/4}=6^{1/4}.$ Hence $\frac{T_5}{\text{5th term from end}}=\left(6^{1/4}\right)^{n-8}=6^{\frac{n-8}{4}}.$

Given this ratio is $\sqrt{6}:1$, i.e. $\frac{T_5}{\text{5th term from end}}=\sqrt{6}=6^{1/2}.$ So, $\frac{n-8}{4}=\frac12$ which gives $n-8=2\quad\Rightarrow\quad n=10.$

5. Now find the third term from the beginning in $\left(\sqrt[4]{2}+\frac{1}{\sqrt[4]{3}}\right)^{10}.$

The third term is $T_3=\binom{10}{2}a^8b^2.$ Compute each part: $\binom{10}{2}=45,$ $a^8=(2^{1/4})^8=2^2=4,$ $b^2=(3^{-1/4})^2=3^{-1/2}=\frac{1}{\sqrt{3}}.$ Therefore, $T_3=45\cdot 4\cdot \frac{1}{\sqrt{3}}=\frac{180}{\sqrt{3}}=60\sqrt{3}.$

6. Therefore the third term from the beginning is $\boxed{60\sqrt{3}}.$ So the correct option is B.