If the middle term in the expansion of ${\left({\mathrm{x}}^2+\frac{1}{\mathrm{x}}\right)}^{\mathrm{n}}$ is $924 {\mathrm{x}}^6$, the value of n is 

Given,

The middle term in the expansion of ${\left(x^2+\frac{1}{x}\right)}^n$ is $924 x^6$

By using binominal expansion, of $(x+y{)}^n$, the $r^{th}$ is of the type,

$T_r{=}^n{C}_{r-1}\left(x{)}^{n-r-1}\right(y{)}^{r-1}$

Since, n is even, therefore ${\left(\frac{n}{2}+1\right)}^{\text{th }}$ term is middle term

Hence, 

${ }^n{C}_{n/2}{\left(x^2\right)}^{n/2}\cdot {\left(\frac{1}{x}\right)}^{n/2}=924x^6$

$$\begin{gathered} \Rightarrow x^{n/2}=x^6 \\ \Rightarrow n=12 \end{gathered}$$

This is the required value.