Total number of terms in the expansion of ${\left(2^{\frac{1}{3}}+\frac{1}{2{\left(3\right)}^{\frac{1}{3}}}\right)}^{10}=10+1=11$

Now, $5^{th}$ term from end $=11-5+1=7^{th}$ from beginning.

We need to find a ratio of the $5^{\mathrm{t}\mathrm{h}}$ term from the beginning to the $5^{\mathrm{t}\mathrm{h}}$ term from the end (i.e. $7^{th}$ term from beginning) in the binomial expansion of ${\left(2^{\frac{1}{3}}+\frac{1}{2{\left(3\right)}^{\frac{1}{3}}}\right)}^{10}=\frac{T_5}{T_7}$

$=\frac{{}^{10}{C}_4{\left(2^{\frac{1}{3}}\right)}^{10-4}{\left(\frac{1}{2{\left(3\right)}^{\frac{1}{3}}}\right)}^4}{{}^{10}{C}_6{\left(2^{\frac{1}{3}}\right)}^{10-6}{\left(\frac{1}{2{\left(3\right)}^{\frac{1}{3}}}\right)}^6} \left[\because T_{r+1}{=}^n{C}_r{\left(a\right)}^{n-r}{\left(b\right)}^r\right]$

$=\frac{{\displaystyle \frac{\left(10!\right)}{\left(4!\right)\left(6!\right)}\times {\left(2\right)}^2\times \frac{1}{\left(2^4\right){\displaystyle {{\displaystyle \left(3\right)}}^{\frac{4}{3}}}}}}{\frac{\left(10!\right)}{\left(6!\right)\left(4!\right)}\times {\left(2\right)}^{{\displaystyle \frac{4}{3}}}\times \frac{1}{\left(2^6\right){\displaystyle {\left(3\right)}^2}}}$

$=4{\left(36\right)}^{\frac{1}{3}}:1$