If 4^th term in the expansion of $(2+3\mathrm{x}{)}^{10}$ is the numerically greatest term, then the value of x can be

$$\begin{gathered} \left|{\mathrm{T}}_4\right|\ge \left|{\mathrm{T}}_3\right|\text{ and }\left|{\mathrm{T}}_4\right|\ge \left|{\mathrm{T}}_5\right|\Rightarrow \\ (2+3\mathrm{x}{)}^{10}=2^{10}{\left(1+\frac{3\mathrm{x}}{2}\right)}^{10},{\mathrm{T}}_4 \mathrm{is} \mathrm{numerically} \mathrm{greatest} \\ \left[\frac{\left(10+1\right)\left|{\displaystyle \frac{3\mathrm{x}}{2}}\right|}{1+\left|\frac{3\mathrm{x}}{2}\right|}\right]=3,4\Rightarrow 3\le \frac{33\left|\mathrm{x}\right|}{2+3\left|\mathrm{x}\right|}⩽4 \\ \Rightarrow \frac{1}{4}\le |\mathrm{x}|\le \frac{8}{21} \end{gathered}$$