Given  expansion  ${(1+x)}^{11}$ ,

According to the problem the fifth term is 24 times the third term

$\begin{array}{c}{T}_5=24T_3\\ \\ \Rightarrow T_{4+1}=24T_{2+1}\\ \\ {\Rightarrow }^{11}{\text{C}}_4{x}^4=24{}^{11}{\text{C}}_2{x}^2\end{array}$ $(\therefore T_{r+1}={}^n{\text{C}}_r{x}^{n-r}{\left(a\right)}^r$be the expansion of ${(x+a)}^n)$

                                                                                   $\left(\therefore {}^n{\text{C}}_r=\frac{n!}{(n-r)!r!}\right)$

$\Rightarrow \frac{11\times 10\times 9\times 8}{1\times 2\times 3\times 4}{x}^4=24\left(\frac{11\times 10}{1\times 2}{x}^2\right)$

$\Rightarrow$ $x^2=4$  

 $\text{therefore \hspace{0.17em}}x=\pm 2$