Given,

The ratio of the fifth term from the beginning to the fifth term from the end in the binomial expansion of ${\left(\sqrt[4]{2}+\frac{1}{\sqrt[4]{3}}\right)}^n$ in the increasing powers of $\frac{1}{\sqrt[4]{3}}$ be $\sqrt[4]{6}:1$

Now according to given data we have,

$\frac{T_5}{T_{n-3}}=\frac{{}^{n}C_4{\left(2^{1/4}\right)}^{n-4}{\left(3^{-1/4}\right)}^4}{{}^{n}C_{n-4}{\left(2^{1/4}\right)}^4{\left(3^{-1/4}\right)}^{n-4}}=\frac{\sqrt[4]{6}}{1}$

$\Rightarrow 2^{\frac{n-8}{4}}{3}^{\frac{n-8}{4}}=6^{1/4}$

$\Rightarrow 6^{n-8}=6$

$\Rightarrow n-8=1\Rightarrow n=9$

So, $T_6={}^{9}C_5{\left(2^{1/4}\right)}^4{\left(3^{-1/4}\right)}^5=\frac{84}{\sqrt[4]{3}}$

Now comparing with $T_6=\frac{\alpha }{\sqrt[4]{3}}$,

We get $\alpha =84$