$\text{ The greatest coefficient in the expansion of }{\left(1+\frac{2x}{3}\right)}^{10}$

$\text{ The greatest coefficient is equal to the greatest term when }x=1$

$\text{ For }x=1\text{ , }$

$\frac{T_{r+1}}{T_r}=\frac{{}^{10}C_r{\left(\frac{2}{3}\right)}^r}{{}^{10}C_{r-1}{\left(\frac{2}{3}\right)}^{r-1}}$

          $=\frac{10-r+1}{r}.\frac{2}{3}$

$\text{ Let }\frac{T_{r+1}}{T_r}\ge 1\Rightarrow \frac{10-r+1}{r}·\frac{2}{3}\ge 1$

$\Rightarrow 20-2r+2\ge 3r$

$\Rightarrow 22\ge 5r$

$\Rightarrow r\le \frac{22}{5}=4.4$

$r=4$

$\text{ Since }r=4,5^{\text{th }}\text{ term is greatest i.e }{T}_5={10}_{c_4}{\left(\frac{2}{3}\right)}^4$