The given expression in the question is ${\left(2-3x^3\right)}^{20}$

${10}^{th}$ term is${ }^{20}{C}_9{\left(-1\right)}^9{\left(2\right)}^{11}{\left(3\right)}^9{x}^{27}$ and ${11}^{th}$ term is ${ }^{20}{C}_{10}{2}^{10}{3}^{10}{x}^{30}$.

Ratio is given, so
$\frac{{ }^{20}{C}_9{\left(-1\right)}^9{\left(2\right)}^{11}{\left(3\right)}^9{x}^{27}}{{ }^{20}{C}_{10}\cdot 2^{10}\cdot 3^{10}\cdot x^{30}}=\frac{45}{22}$
$\Rightarrow -\frac{10}{11}\cdot \frac{2}{3x^3}=\frac{45}{22}$
$\Rightarrow x^3=-\frac{8}{27}\Rightarrow x=-\frac{2}{3}$