This problem comes from Jech exercise 12.11:
If $\kappa$ is an inaccessible cardinal then $V_\kappa \models$ there is a countable model of ZFC.
I see this question partially answered here: Jech, "Set theory" exercises 12.11 - Is my proof right?
But I still have a question. What does it mean for $V_\kappa \models (\mathfrak{U} \models $ ZFC) ? If my impression is right only a "$\sigma$" should follow the "$\models$" where $\sigma$ is a sentence or formula in the language of set theory. How do you turn "$\mathfrak{U} \models $ ZFC" into a sentence in the language set theory.
In the post linked above the author only showed that $(\varphi^{\omega,E})^{V_\kappa}$ held for each axiom of ZFC, but is that the same thing as showing (or all that is required to show) $V_\kappa \models (\omega, E)$ is a model of ZFC ?
Fix a transitive model $\textbf{K}$ of ZF (possibly with $\mathbf{K}$ a proper class).
When I say that a statement $\Phi(a_1, ..., a_n)$ in the language of set theory is absolute, I mean that $\forall a_1,\in \textbf{K} \ldots \forall a_n \in \textbf{K}, \Phi^{\textbf{K}}(a_1, \ldots, a_n) \iff \Phi(a_1, \ldots, a_n)$. Here, $\Phi^{\textbf{K}}(a_1, \ldots, a_n)$ is the statement $\Phi(a_1, \ldots, a_n)$ with all quantifiers relativised to $\mathbf{K}$.
First of all, given a vocabulary $V$ and a statement $\phi(x_1, ..., x_n)$ in the language $V$, and given a structure $M$ and values $a_1, ..., a_n \in M$, we know how to define $M, x_1 \gets a_1, ..., x_n \gets a_n \models \phi(x_1, ..., x_n)$. In particular, this is done by induction on $\phi(x_1, ..., x_n)$. Note that here, $V$, $\phi(\ldots)$, and $M$ are objects within set theory.
The key here is to verify that the statement $M, x_1 \gets a_1, ..., x_n \gets a_n \models \phi(x_1, ...., x_n)$ is absolute for all statements $\phi$ in the vocabulary in any transitive model of ZF. This is easy to prove, since the statement involves only $\Delta_0$ quantifiers.
Given a set $S$ of sentences, we say a structure $M$ models $S$ if and only if $\forall s \in S, M \models S$. Once again, this statement involves only a $\Delta_0$ quantifier, so it is absolute.
We also know how to define the vocabulary $K$ of ZFC. It's just $V = \{\in\}$. And we know how to define the set $S$ of axioms of ZFC. You should verify that both $V$ and $S$ are absolute. This is pretty straightforward, but a bit tedious.
Therefore, the statement "$M$ models ZFC" is absolute.
Now suppose that there is an inaccessible cardinal $\kappa$. Then ZFC is consistent. Then by the downward Lowenheim-Skolem theorem, there is a countable model $M$ of ZFC. Since $M$ is countable, we can place it into bijection with $\mathbb{N}$, and thus we can obtain a model $(\mathbb{N}, \in_\mathbb{N})$ of ZFC.
Now $\mathbb{N} \in V_\kappa$, and also $P(\mathbb{N}) \in V_\kappa$. Since $V_\kappa$ is transitive, $\in_\mathbb{N} \in V_\kappa$. Therefore, $(\mathbb{N}, \in_\mathbb{N}) \in V_\kappa$.
Since $V_\kappa$ is a transitive model of ZFC, we see that the statement $(\mathbb{N}, \in_\mathbb{N}) \models ZFC$ is absolute for $V_\kappa$. Since the statement is true, $((\mathbb{N}, \in_\mathbb{N}) \models ZFC)^{V_\kappa}$ holds. And of course, $V_\kappa \models \mathbb{N}$ is countable.
Therefore, we see that $V_\kappa \models$ there exists a countable model of ZFC.
Note that therefore, it's actually the case that if there is an accessible cardinal, then ZFC + ("ZFC is consistent") is consistent. This means that there is a countable model of ZFC + ("ZFC is consistent"). And therefore, $V_\kappa \models$ there is a countable model of ZFC + ("ZFC is consistent"). Therefore, ZFC + "ZFC is consistent" + "(ZFC + ("ZFC is consistent")) is consistent". We can repeat this ad infinitum.
So the existence of an inaccessible cardinal is much, much stronger than the consistency of ZFC.