The smallest natural numerb ‘n’ such that the coefficient of ‘x’ in the expansion of ${\left[x^2+\frac{{\displaystyle 1}}{{\displaystyle x^3}}\right]}^n$ is $n_{C_{23}}$ is 

$T_{r+1}{=}^n{C}_r{\left(x\right)}^{2n-5r}$ for the coefficient of X.
$2n-5r=1\Rightarrow 2n=5r+1\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(i\right)$
As coefficient of  x is given as $\text{'}\text{'}{C}_{23},$ hence either 
$r=23 \text{\hspace{0.17em}\hspace{0.17em}}or \text{\hspace{0.17em}}n-r=23$
If r=23, then from eq (i)  we get
$$\begin{gathered} 2n=5\left(23\right)+1 \\ 2n=115+1 \\ \Rightarrow 2n=116\Rightarrow n=58 \end{gathered}$$
If n-r=23, then from eq (i) n=38
So, the required smallest natural number n=38