Let $$\alpha > 0$$, be the smallest number such that the expansion of $$\left(x^{\frac{2}{3}}+\frac{2}{x^3}\right)^{30}$$ has a term $$\beta x^{-\alpha}$$, $$\beta \in N$$. Then $$\alpha$$ is equal to ______.

We need to find the smallest $$\alpha > 0$$ such that the expansion of $$\left(x^{2/3} + \frac{2}{x^3}\right)^{30}$$ has a term $$\beta x^{-\alpha}$$, $$\beta \in \mathbb{N}$$.

$$T_{r+1} = \binom{30}{r} \left(x^{2/3}\right)^{30-r} \left(\frac{2}{x^3}\right)^r = \binom{30}{r} 2^r x^{\frac{2(30-r)}{3} - 3r}$$

Power of $$x$$: $$\frac{2(30-r)}{3} - 3r = \frac{60 - 2r - 9r}{3} = \frac{60 - 11r}{3}$$

$$\frac{60 - 11r}{3} = -\alpha \Rightarrow \alpha = \frac{11r - 60}{3}$$

For $$\alpha > 0$$: $$11r > 60 \Rightarrow r > \frac{60}{11} \approx 5.45$$, so $$r \geq 6$$.

For $$\alpha$$ to be a positive integer (or positive value with $$\beta \in \mathbb{N}$$), we need $$11r - 60$$ divisible by 3.

$$11r - 60 \equiv 2r \pmod{3}$$. So we need $$r \equiv 0 \pmod{3}$$.

Smallest valid $$r$$: $$r = 6$$ (since $$6 > 5.45$$ and $$6 \equiv 0 \pmod{3}$$).

$$\alpha = \frac{66 - 60}{3} = 2$$

Check $$\beta$$: $$\beta = \binom{30}{6} \cdot 2^6$$, which is a natural number. ✓

The answer is $$\boxed{2}$$.