$T_5$ in $\big[2^{\Large\frac{1}{4}}+\large\frac{1}{3^{\Large\frac{1}{4}}}\big]^n$$=nC_4(2^{1/4})^{n-4}.\big(\large\frac{1}{3^{1/4}}\big)^4$
$\Rightarrow nC_42^{\Large\frac{n-4}{4}}.\large\frac{1}{3}$-------(1)
Total number of terms =n+1
Fifth term from the end =$[(n+1)-5+1)]^{th}$ term from the beginning.
$\Rightarrow (n-3)^{th}$ term from the beginning.
$\Rightarrow nC_{n-4}(2^{1/4})^{n-(n-4)}\big(\large\frac{1}{3^{1/4}}\big)^{n-4}$
$\Rightarrow nC_4.2(\large\frac{1}{3})^{\Large\frac{n-4}{4}}$------(2)
Dividing (1) by (2)
$\large\frac{nC_42^{\Large\frac{n-4}{4}}.\large\frac{1}{3}}{nC_4.2(\large\frac{1}{3})^{\Large\frac{n-4}{4}}}=\frac{\sqrt 6}{1}$
Or $\large\frac{2^{\Large\frac{n}{4}-2}}{(1/3)^{\Large\frac{n}{4}-2}}=\frac{\sqrt 6}{1}$
Or $2^{\Large\frac{n}{4}-2}.3^{\Large\frac{n}{4}-2}=6^{\Large\frac{1}{2}}$
$6^{\Large\frac{n}{4}-2}=6^{\Large\frac{1}{2}}$
$\therefore \large\frac{n}{4}$$-2=\large\frac{1}{2}$
$\therefore \large\frac{n}{4}=\frac{5}{2}$
$n=\large\frac{5}{2}$$\times 4$
$n=10$
Hence (C) is the correct answer.