lf the term independent of x in the expansion of ${\left(\frac{3}{2}{x}^2-\frac{1}{3x}\right)}^9$is k, then 18k is equal to

given binomial is ${\left(\frac{3}{2}{x}^2-\frac{1}{3x}\right)}^9$So the general term,

$\begin{array}{c}{\mathrm{T}}_{\mathrm{r}+1}{=}^9{\mathrm{C}}_{\mathrm{r}}{\left(\frac{3}{2}{\mathrm{x}}^2\right)}^{9-\mathrm{r}}{\left(-\frac{1}{3\mathrm{x}}\right)}^{\mathrm{r}}\\ {=}^9{\mathrm{C}}_{\mathrm{r}}{\left(\frac{3}{2}\right)}^{9-\mathrm{r}}{\left(-\frac{1}{3}\right)}^{\mathrm{r}}{\mathrm{x}}^{18-3\mathrm{r}}\end{array}$

lf the term is independent of x, then $18-3\mathrm{r}=0\Rightarrow \mathrm{r}=6$

$\therefore (\mathrm{r}+1)\text{ th }$ term =7th term is independent of x

Now, as ${\mathrm{T}}_{6+1}=\mathrm{k}$

$\begin{array}{l}\Rightarrow {}^9{\mathrm{C}}_6{\left(\frac{3}{2}\right)}^3{\left(-\frac{1}{3}\right)}^6=\mathrm{k}\\ \Rightarrow \frac{9\times 8\times 7}{3\times 2}\times \frac{1}{27\times 8}=\mathrm{k}\Rightarrow \mathrm{k}=\frac{7}{18}\Rightarrow 18\mathrm{k}=7\end{array}$

Hence, option (b) is correct.