As I am studying set theory, I came to realize that there exists a countable "well-founded" model of ZFC. But I am curious whether countable models can ever be well-founded externally. What would be the way to see whether this is the case or not?
When working in set theory, we have a universe of sets, $V$ with an $\in$ relationship. And we work inside of that universe. Since this universe is usually assumed to satisfy the axioms of $\sf ZFC$ this means that $V$ is well-founded, because as a universe of sets, it is the class of all sets.
When we say that $(M,E)$ is a model of $\sf ZFC$ then we say that $M$ and $E$ are sets in $V$ and that $V$ thinks that all the axioms of $\sf ZFC$ are true in $(M,E)$. It should be pointed that $V$ may have "its own version" of $\sf ZFC$ which differs than ours, but let's deal with one obstacle at a time.
When we say that $(M,E)$ is a well-founded model, then we mean that $E$ is a well-founded relation. When we say that $M$ is a transitive model, then we mean that $E=\in$. That is, the membership relation is the same as that of $V$. Since $V$ satisfies the axiom of foundation, this means that $\in$ is well-founded, so $M$ is well-founded.
It should be pointed that every well-founded model of $\sf ZFC$ is isomorphic to a transitive one (and the isomorphism is unique, too). This is by the Mostowski collapse lemma.