Solution: We will use Rouché's theorem to solve this problem. Let us recall the Rouché's theorem.
(Rouché Theorem) Let $f,g: U \to \mathbb{C} $ be analytic on $U \subset \mathbb{C} $ open, and let $\gamma $ be any closed curve such that $\gamma \subset U$. If $\vert f(z) \vert \leq \vert g(z) \vert $ on $\gamma $ then $f(z)$ and $f(z) = g(z)$ have the same number of zeros inside $\gamma $.
Let $f(z) = z^5 + 6z^3 - 10$, $g(z) = z^5$ and $h(z) = 6z^3$. By the triangle's inequality and $\vert z \vert = 3$, we have
\[
\vert f(z) - g(z) \vert = \vert 6z^3 - 10 \vert \leq 6 \vert z \vert ^3 + 10 = 226.
\]
Also, if $\vert z \vert \lt 3$, then $\vert g(z) \vert = 729$. Thus, for $\vert z \vert = 3$,
\[
\vert f(z) - g(z) \vert \lt \vert g(z) \vert .
\]
Similarly, on the circle $\vert z \vert = 2$, we have
\[
\vert f(z) - h(z) \vert \lt h(z).
\]
So, by Rouché's theorem, $f$ has $5$ zeros (counting with multiplicities) inside the open disk $\vert z \lt 3 \vert $ and three zeros (counting with multiplicities) inside the open disk $\vert z \vert \lt 2$. Since $f$ has no zeros on the smaller circle $\vert z \vert = 2$, (by the triangle inequality $\vert f(z) \vert \lt \vert g(z) \vert $ ). Thus, the total number of zeros of $f$ inside the annulus is $5 - 3 = 2$.