1. Write the general term

For $\left(\frac{3}{2}x^2 - \frac{1}{3x}\right)^9,$ the general term is $T_{r+1} = {9 \choose r}\left(\frac{3}{2}x^2\right)^{9-r}\left(-\frac{1}{3x}\right)^r.$

2. Simplify the power of $x$

The power of $x$ in this term is $x^{2(9-r)}\cdot x^{-r} = x^{18-2r-r} = x^{18-3r}.$

For the term independent of $x$, we need $18-3r=0 \implies r=6.$

So the constant term is $T_7$.

3. Compute the constant term

Substitute $r=6$: $k = {9 \choose 6}\left(\frac{3}{2}\right)^3\left(-\frac{1}{3}\right)^6.$

Now, $\binom{9}{6}=84,$ $\left(\frac{3}{2}\right)^3=\frac{27}{8},$ $\left(-\frac{1}{3}\right)^6=\frac{1}{729}.$

Thus, $k=84\cdot \frac{27}{8}\cdot \frac{1}{729}.$

Simplify: $84\cdot \frac{27}{8} = \frac{84}{8}\cdot 27 = \frac{21}{2}\cdot 27 = \frac{567}{2},$ so $k=\frac{567}{2}\cdot \frac{1}{729} = \frac{567}{1458} = \frac{7}{18}.$

4. Find $18k$

$18k = 18\cdot \frac{7}{18} = 7.$

5. Match with options

So, $18k=7,$ which corresponds to Option C.