Let $$m$$ and $$n$$ be the coefficients of seventh and thirteenth terms respectively in the expansion of $$\left(\frac{1}{3}x^{1/3} + \frac{1}{2x^{2/3}}\right)^{18}$$. Then $$\left(\frac{n}{m}\right)^{1/3}$$ is:

The binomial expansion is given by: $$\left(\frac{1}{3}x^{1/3} + \frac{1}{2x^{2/3}}\right)^{18}$$.

The general term in the expansion of $$(a + b)^k$$ is $$T_{r+1} = \binom{k}{r} a^{k-r} b^r$$.

Here, $$a = \frac{1}{3}x^{1/3}$$, $$b = \frac{1}{2} x^{-2/3}$$, and $$k = 18$$.

So, the general term is: $$T_{r+1} = \binom{18}{r} \left( \frac{1}{3} x^{1/3} \right)^{18-r} \left( \frac{1}{2} x^{-2/3} \right)^r$$

Simplifying: $$T_{r+1} = \binom{18}{r} \left( \frac{1}{3} \right)^{18-r} (x^{1/3})^{18-r} \cdot \left( \frac{1}{2} \right)^r (x^{-2/3})^r = \binom{18}{r} \left( \frac{1}{3} \right)^{18-r} \left( \frac{1}{2} \right)^r x^{\frac{18-r}{3} - \frac{2r}{3}}$$

The exponent of $$x$$ is: $$\frac{18-r}{3} - \frac{2r}{3} = \frac{18 - r - 2r}{3} = \frac{18 - 3r}{3} = 6 - r$$

Thus: $$T_{r+1} = \binom{18}{r} \left( \frac{1}{3} \right)^{18-r} \left( \frac{1}{2} \right)^r x^{6 - r}$$

The coefficient of $$T_{r+1}$$ is: $$\binom{18}{r} \left( \frac{1}{3} \right)^{18-r} \left( \frac{1}{2} \right)^r$$

The seventh term corresponds to $$r+1 = 7$$, so $$r = 6$$.

The thirteenth term corresponds to $$r+1 = 13$$, so $$r = 12$$.

Let $$m$$ be the coefficient of the seventh term and $$n$$ be the coefficient of the thirteenth term.

So: $$m = \binom{18}{6} \left( \frac{1}{3} \right)^{12} \left( \frac{1}{2} \right)^6$$ $$n = \binom{18}{12} \left( \frac{1}{3} \right)^{6} \left( \frac{1}{2} \right)^{12}$$

Since $$\binom{18}{12} = \binom{18}{6}$$, we have: $$n = \binom{18}{6} \left( \frac{1}{3} \right)^{6} \left( \frac{1}{2} \right)^{12}$$

Now, compute $$\frac{n}{m}$$: $$\frac{n}{m} = \frac{ \binom{18}{6} \left( \frac{1}{3} \right)^{6} \left( \frac{1}{2} \right)^{12} }{ \binom{18}{6} \left( \frac{1}{3} \right)^{12} \left( \frac{1}{2} \right)^{6} } = \frac{ \left( \frac{1}{3} \right)^{6} \left( \frac{1}{2} \right)^{12} }{ \left( \frac{1}{3} \right)^{12} \left( \frac{1}{2} \right)^{6} } = \left( \frac{1}{3} \right)^{6-12} \left( \frac{1}{2} \right)^{12-6} = \left( \frac{1}{3} \right)^{-6} \left( \frac{1}{2} \right)^{6} = 3^{6} \cdot 2^{-6} = \left( \frac{3}{2} \right)^6$$

Now, compute $$\left( \frac{n}{m} \right)^{1/3}$$: $$\left( \frac{n}{m} \right)^{1/3} = \left( \left( \frac{3}{2} \right)^6 \right)^{1/3} = \left( \frac{3}{2} \right)^{6 \cdot \frac{1}{3}} = \left( \frac{3}{2} \right)^2 = \frac{9}{4}$$

The value is $$\frac{9}{4}$$, which corresponds to option D.