The answer is no, because I claim that if $\kappa$ is worldly, then $V_\kappa$ thinks that every set is a member of a transitive model of ZFC.

To see this, note first that every worldly cardinal $\kappa$ is a beth-fixed point $\beth_\kappa=\kappa$ and furthermore $V_\kappa=H_\kappa$, the set of sets whose transitive closures have size less than $\kappa$. Now consider any $x\in V_\kappa$. By the Löwenheim-Skolem theorem, we can find an elementary substructure $X\prec V_\kappa$ with $x\subseteq X$ and $x\in X$, with $|X|=|\text{TC}(x)|<\kappa$. The transitive collapse $M$ of $X$ will be a model of ZFC containing $x$ as an element. And even though $\kappa$ is singular, and so perhaps $X$ is unbounded in $V_\kappa$, nevertheless we will have $M\in V_\kappa$ since it is small enough. For example, $M$ will have fewer than $\kappa$ many ordinals, and so $M\subseteq V_\beta$ for some $\beta<\kappa$ and hence $M\in V_{\beta+1}\subset V_\kappa$.

Incidentally, this axiom, that every set is an element of a transitive model of ZFC, has sometimes been put forth as an alternative to the Grothendieck universe axiom, asserting that there are unboundedly many inaccessible cardinals, since it captures much of the power of that axiom in its applications, by providing a robust small universe concept for any given set, while having considerably weaker large cardinal strength. But to use this axiom, one needs to make the move from Grothendieck universes to transitive models of ZFC, which do not necessarily compute the power set correctly and whose height may be singular, although this is not visible internally to the model.

The axiom has also been proposed as a natural arena in which to investigate the corresponding multiverse theory, a version of hyperversism, but with uncoutnable models. For this reason, I like the axiom very much. In some recent upcoming joint work with Øysten Linnebo, we analyze the modal logic of this collection of models, viewed as a Kripke model.

mayfind your answer in the "Transitive $ZFC$ Model" entry of Cantor's Attic (it's in the first paragraph of that entry). Does the argument contained therein answer your question? $\endgroup$ – Thomas Benjamin Apr 28 '17 at 8:00