A model of ZFC is a set $E$ together with a binary relation $R$ satisfying the axioms of ZFC (and we can always suppose that $E$ is transitive and that $R=\in|_{E\times E}$). So we not only need to find a class of sets satisfying the axioms of ZFC, but this class must be itself a set.

If you have a strongly inaccessible cardinal $\kappa$, then there is a set $V_\kappa$ which is a model of ZFC, hence ZFC is consistent. But without the existence of $\kappa$, even if you could consider the class of all sets of rank “less than $\kappa$”, this would not be a set and the theorem ”there is a model implies the consistency of ZFC” would not be applicable (because there is no model).

A model of ZFC is a set $E$ together with a binary relation $R$ satisfying the axioms of ZFC (and we can always if we can suppose that $E$ is transitive and that $R=\in|_{E\times E}$ then the model is said standard). So we not only need to find a class of sets satisfying the axioms of ZFC, but this class must be itself a set.

If you have a strongly inaccessible cardinal $\kappa$, then there is a set $V_\kappa$ which is a (standard) model of ZFC, hence ZFC is consistent. But without the existence of $\kappa$, even if you could consider the class of all sets of rank “less than $\kappa$”, this would not be a set and the theorem ”there is a model implies the consistency of ZFC” would not be applicable (because there is no model).