This is exercise II.5.6 from Kunen's Set Theory:
Assume Axiom of Choice. Find a formula $\varphi$ such that every transitive $M$ satisfying $M\preccurlyeq _\varphi V$ is of the form $R(\gamma)$ for some $\gamma = \beth _\gamma$.
(These $R(\gamma)$ are the "approximations of the universe" sometimes noted $V_\gamma$.)
It includes a hint:
$\varphi$ can be $\varphi_{0}(x)$$\land$$\varphi_{1}$ where $\varphi_0(x)$ says that $x$ is the set of the form $R(\alpha)$, and $\varphi_1$ is a sentence and a theorem of $ZFC$.
I imagine Kunen meant to write "where $\varphi_0(x)$ says that $x$ is $\textbf{a}$ set of the form $R(\alpha)$" but I copy it here verbatim just in case.
I don't get this hint, it seems to say $\varphi_1$ can be any theorem of $ZFC$ but this doesn't make sense as taking $\varphi_1$ to be $\exists z (z=z)$ we get $R(\omega)\preccurlyeq_\varphi V$ as for every $x\in R(\omega)$ we have $\varphi_0(x) ^{R(\omega)} \iff \varphi_0(x) ^{V}$. But $\omega \neq \beth_\omega$.
Am I misinterpreting the hint? I need help solving this exercise. Thank you for reading.
P.S. This exercise is probably related to the Reflection Theorem as it is given in the chapter dedicated to it.
The point is that if $\varphi=\varphi_0(x)\wedge\varphi_1$ for $\varphi_1$ a theorem of ZFC, then $M\preccurlyeq_\varphi V$ iff
$M\preccurlyeq_{\varphi_0}V$, and
$M\models \varphi_1$.
So we're really getting to constrain $M$ in two ways: $M$ will be correct about $\varphi_0$, and $M$ will satisfy $\varphi_1$.
For example, take $\varphi_0(x)=$"$x=\langle y, z\rangle$ for some $y, z$ such that $y=\mathcal{P}(z)$" and $\varphi_1$ to be a large finite fragment of ZFC containing the Powerset axiom. Then $M\preccurlyeq_{\varphi}V$ implies that $M$ is closed under powersets: the $\varphi_1$-part says that $M$ contains "internal powersets" - that is, for each $z\in M$ there is some $y\in M$ such that $M$ thinks $y$ is the powerset of $z$ - and the $\varphi_0$-part implies that the "internal powersets" are in fact actual powersets.
Does this help?