If the middle term in the binomial expansion of ${(\frac{1}{x}+x\sin x)}^{10}$  is $\frac{63}{8},$  then the
value of $6{\sin }^{2}x+\sin x-2$  is

         Here, n = 10, which is even.
Middle term $={(\frac{10}{2}+1)}^{th}$  term = $6^{th}$  term
 $T_6{=}^{10}{C}_5{\left(\frac{1}{x}\right)}^5{\left(x\sin x\right)}^5$

$\Rightarrow \frac{63}{8}=252{\left(\sin x\right)}^5$
 $\Rightarrow {\left(\sin x\right)}^5=\frac{1}{32}$
 
 $\Rightarrow {\left(\sin x\right)}^5={\left(\frac{1}{2}\right)}^5$
 $\Rightarrow \sin x=\frac{1}{2}$
 $\Rightarrow 2\sin x-1=0$
 $\Rightarrow 6{\sin }^{2}x+\sin x-2=(2\sin x-1)(3\sin x+2)=0$