Answer: In the binomial expansion of $(x^2 + 2/x)^{13}$, the 13th term will be independent of $x$ if the exponents of $x$ in the numerator and denominator cancel each other out. In other words, the 13th term will have the form
$$\binom{13}{n} x^{13 - 2n} (2/x)^n = \frac{(13!)}{(13 - n)!n!} x^{13 - 2n} (2^n/x^n)$$
For this term to be independent of $x$, we must have $13 - 2n = 0$, which means $n = \boxed{6}$. This means that the 13th term in the binomial expansion of $(x^2 + 2/x)^{13}$ is given by
$$\binom{13}{6} x^{13 - 2(6)} (2/x)^6 = \frac{(13!)}{(13 - 6)!6!} x^{13 - 2(6)} (2^6/x^6) = \frac{13!}{7!6!} (2^6) = \frac{13!}{7!6!} 64$$
which does not depend on the value of $x$.