$\left|{\mathrm{T}}_3\right|<\left|{\mathrm{T}}_4\right|>\left|{\mathrm{T}}_5\right|$
$\Rightarrow \left|\frac{{\mathrm{T}}_4}{{\mathrm{T}}_3}\right|>1$ and $\left|\frac{{\mathrm{T}}_5}{{\mathrm{T}}_4}\right|<1$
${\mathrm{T}}_{\mathrm{r}+1}={}_{\mathrm{}}{}^{10}{\mathrm{C}}_r{2}^{10-r}{\left(\frac{3x}{8}\right)}^r$
$\frac{{\text{T}}_4}{{\text{T}}_3}=\frac{{}_{}{}^{10}\text{C}{}_3{2}^7{\left(\frac{3x}{8}\right)}^3}{{}_{}{}^{10}\text{C}{}_2{2}^8{\left(\frac{3x}{8}\right)}^2}=\frac{x}{2}$
$\frac{{\text{T}}_5}{{\text{T}}_4}=\frac{{}_{}{}^{10}\text{C}{}_4{2}^6{\left(\frac{3x}{8}\right)}^4}{{}_{}{}^{10}\text{C}{}_3{2}^7{\left(\frac{3x}{8}\right)}^3}=\frac{21}{64}x$
$\left|\frac{x}{2}\right|>1\Rightarrow x>2$ or $x<-2$
$\left|\frac{21}{64}x\right|<1\Rightarrow -\frac{64}{21}<x<\frac{64}{21}$
Taking intersection, we get,

 $x\in \left(-\frac{64}{21},-2\right)\cup \left(2,\frac{64}{21}\right)$

Hence, integral values of $x=-3,\mathrm{}+3$