Let $\langle M,E\rangle$ be a model of $\mathsf{ZFC}$.

Does there exist a $d\in M$ such that, for all $a\in M$, $a\mathrel{E}d$ if and only if $a$ is definable in $\langle M,E\rangle$ without parameters?

A result of J.D. Hamkins, etc. (cf. Pointwise definable models of set theory) shows that, for some models of $\mathsf{ZFC}$, such a $d$ cannot exist. Is there a model in which such a $d$ exists?

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    $\begingroup$ If $\kappa$ is an inaccessible cardinal, then $(V_\kappa,\in)$ is a model of ZFC in which there is a set $d$ of all elements that are definable without parameters. Since there are only countably many definable elements, their ranks are bounded below $\kappa$, and any subset of $V_\kappa$ with rank below $\kappa$ is an element of $V_\kappa$. $\endgroup$ – Andreas Blass Mar 5 at 3:58
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    $\begingroup$ And by the condensation property, the $V_{\kappa}$ can be replaced with $L_{\kappa}$ in Prof. Blass's answer. Also $\kappa$ can be just inaccessible in $L$. $\endgroup$ – 喻 良 Mar 5 at 5:29
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    $\begingroup$ Andreas's argument is example (v) on page 146 of the article cited in the OP. $\endgroup$ – Joel David Hamkins Mar 5 at 13:25

In the paper you mention in the original post, we mention several of the possibilities as follows. Item (v) includes the particular situation you asked about.

Hamkins, Joel David; Linetsky, David; Reitz, Jonas, Pointwise definable models of set theory, J. Symb. Log. 78, No. 1, 139-156 (2013). ZBL1270.03101. blog post

Let us now turn to the question of the extent to which definability is first-order expressible, by presenting a number of examples that illustrate the range of possibility. We have already observed that the property of a model being pointwise definable is not first order expressible, since it is not preserved by nontrivial elementary extensions. Since pointwise definability is a strong generalization of the axiom $V=\newcommand\HOD{\text{HOD}}\HOD$, it is tempting to introduce such notation as $V=D$ or $V=HD$ to express that a model is pointwise definable, thereby maintaining a parallel to the classical $V=\HOD$ notation while emphasizing that the definitions need no parameters. We hesitate to adopt this notation, however, because we fear it would incorrectly suggest that the concept is first-order expressible, which isn't the case.

(i) There is no uniform definition of the class of definable elements. Specifically, there is no formula $\mathop{\rm df}(x)$ in the language of set theory that is satisfied in any model $M\newcommand\satisfies{\models}\satisfies\newcommand\ZFC{\text{ZFC}}\ZFC$ exactly by the definable elements. The reason is that if $M_0$ is pointwise definable and $M_0\prec M$ is a nontrivial elementary extension, then the definable elements of $M_0$ and $M$ are precisely the elements of $M_0$, and so $M_0$ should satisfy $\forall x\,\mathop{\rm df}(x)$ but $M$ would satisfy $\exists x\,\neg \mathop{\rm df}(x)$, contrary to $M_0\prec M$.

(ii) The class of definable elements can form a definable class. Although there is no uniform definition of the class of definable elements, it can sometimes happen that a model enjoys a certain structure that allows it to see its collection of definable elements as a definable class. For example, in a pointwise definable model, the class of definable elements includes every object and is therefore defined by the formula $x=x$. See also (iv) and (v) below.

(iii) The collection of definable elements might not form a class. Consider any model $M\satisfies\ZFC$, and let $N$ be an ultrapower of $M$ by an ultrafilter on the natural numbers. The parameter-free definable elements of $N$ are necessarily contained in the range of the ultrapower map, and in particular, do not include any of the newly added nonstandard natural numbers. Thus, the class of definable elements of $N$ is not amenable to $N$, for it would reveal that its natural number are not well-founded.

(iv) The definable elements can form a definable class in a model having no class function $r\mapsto\psi_r$ mapping definable elements to definitions. Suppose that $M$ is a pointwise definable model of $\ZFC$. The definable elements of $M$ are all of $M$, which is certainly a definable class in $M$. But $M$ cannot have a function $r\mapsto\psi_r$ associating to each element $r$ of $M$, or even to each real of $M$, a defining formula $\psi_r$, since such a map would reveal to $M$ that it has only countably many reals.

(v) The definable elements can be a set in a model that does have a definability map $r\mapsto\psi_r$. Suppose that $\kappa$ is an inaccessible cardinal (this hypothesis can be reduced), and observe by a Lowenheim-Skolem argument that there are numerous $\gamma<\kappa$ with $V_\gamma\prec V_\kappa\satisfies\ZFC$. It follows that the definable elements of $V_\kappa$ are all in $V_\gamma$ and satisfy the same definitions there as in $V_\kappa$. Since $V_\gamma$ is a set in $V_\kappa$, we may construct in $V_\kappa$ the function $r\mapsto \psi_r$ that maps every definable element $r$ of $V_\gamma$ to the smallest definition $\psi_r$ of it, and because $V_\gamma\prec V_\kappa$, this function has the same property with respect to $V_\kappa$, as desired. The large cardinal hypothesis can be reduced; it is sufficient to have an $\omega$-model $M$ with some $M_0\in M$ having $M_0\prec M$.

(vi) No model can have a definable definability map $r\mapsto\psi_r$. If such a map were definable, then since there are only countably many definitions $\psi_r$, we could easily diagonalize against it to produce a definable real not in the domain of the map. In (v), the map is definable from parameter $\gamma$.

The surviving content of the math-tea argument seems to be the observation that in any model with access to a definability map $r\mapsto\psi_r$, the definable reals do not exhaust all the reals.

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    $\begingroup$ Talk about covering all bases... $\endgroup$ – Asaf Karagila Mar 6 at 7:41

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