1. Write the general term

For $\left(x^{\frac{2}{3}}+\frac{2}{x^3}\right)^{30},$ the general term is $T_{r+1}=\binom{30}{r}\left(x^{\frac{2}{3}}\right)^{30-r}\left(\frac{2}{x^3}\right)^r,$ where $r=0,1,2,\dots,30$.

2. Simplify the power of $x$

$T_{r+1}=\binom{30}{r}2^r x^{\frac{2}{3}(30-r)}x^{-3r}.$ So the exponent of $x$ is $\frac{2}{3}(30-r)-3r = 20-\frac{2r}{3}-3r = 20-\frac{11r}{3}.$

Thus the general term is $T_{r+1}=\binom{30}{r}2^r x^{\,20-\frac{11r}{3}}.$

3. We want a term of the form $\beta x^{-\alpha}, \quad \beta\in\mathbb N,$ with $\alpha>0$ smallest.

That means we want the exponent $20-\frac{11r}{3}$ to be negative, but as close to $0$ as possible.

Let $20-\frac{11r}{3}=-\alpha.$ Then $\alpha=\frac{11r}{3}-20.$ We need $\alpha>0$, so $\frac{11r}{3}>20 \implies 11r>60 \implies r>\frac{60}{11}.$ Hence the smallest integer $r$ is $r=6.$

4. Compute the corresponding exponent

For $r=6$, $20-\frac{11\cdot 6}{3}=20-22=-2.$ So the term is of the form $\beta x^{-2},$ where $\beta=\binom{30}{6}2^6 \in \mathbb N.$ Thus, $\alpha=2.$

5. Check smaller values

For $r=5$, $20-\frac{55}{3}=\frac{5}{3}>0,$ so it is not negative. Hence $r=6$ indeed gives the smallest positive $\alpha$.

Therefore, $\boxed{\alpha=2}.$

6. Comparison with stored answer

Stored correct answer = $2$.

Our derived answer also equals $2$, so they agree.