In the expansion of ${\left(1+\frac{3x}{2}\right)}^{-5},$ the coefficient of $x^{10}$ is equal to the coefficient of $x^{10}$ in ${\left(1+ax\right)}^n$.

General term of expansion

$=\frac{\left(-5\right)\left(-5-1\right)\left(-5-2\right)......\left(-5-10+1\right)}{10!}\times {\left(\frac{3x}{2}\right)}^{10}$

$=\frac{\left(-5\right)\left(-6\right)\left(-7\right)......\left(-14\right)}{10!}\times \frac{3^{10}}{2^{10}}{x}^{10} ......\left(i\right)$

Coefficient of $x^{10}$ in ${\left(1+ax\right)}^n$

$T_{10+1}={}^n{C}_{10} {\left(ax\right)}^{10} ......\left(ii\right)$

Equation $\left(i\right)\&\left(ii\right)$ both are equal

$\frac{\left(-5\right)\left(-6\right)\left(-7\right)......\left(-14\right)}{10!}\times \frac{3^{10}}{2^{10}}{x}^{10}={}^n{C}_{10} {\left(ax\right)}^{10}$

$n=7, a=3$

Hence, the value of $na=21$.