 Solution

False. Use the two-generated group $G_S$ constructed in part (iii) for a non-recursively-enumerable $S$. Sending the two free generators to $b_0,t$ gives a surjective group homomorphismThe source is a finitely presented group, with presentation $⟨a,b\mid ⟩$. The target is not finitely presented: if it were, its null words could be enumerated from a finite set of relators, and the tested commutators would enumerate $S$, as proved in part (iii). Thus even a quotient of a rank-two free group need not admit a finite presentation. A quotient by a finitely normally generated kernel would be finitely presented, but the assertion places no such restriction on its kernel.