Numerically greatest term in the expansion of ${(2x-3y)}^{11}$ when $x=\frac{1}{3}$ and $y=\frac{1}{2}$ is 

${(2x-3y)}^{11}={\left(2x\right)}^{11}{(1-\frac{3y}{2x})}^{11}={\left(2x\right)}^{11}{(1+\alpha )}^{11}$

Where $\alpha =\frac{-3y}{2x}=\frac{-9}{4}$

Now $\frac{(n+1)\left|\alpha \right|}{\left|\alpha \right|+1}=\frac{12\times \frac{9}{4}}{\frac{9}{4}+1}=\frac{108}{13}=8.3i.e{9}^{th}term$

$\therefore T_9={11}_{C_8}{\left(2x\right)}^3{(-3y)}^8={11}_{C_3}{\left(\frac{3}{2}\right)}^5$