Let $\kappa$ be the supremum of ordinals first order definable in L without parameters. Assume $0^\sharp$ exists. Is $\kappa$ the least silver indiscernible ordinal?
Updated. The answer is no.
First, let me point out that in general, in ZFC we are not able to refer to the notion of firstorderdefinablein$L$, since definability is not expressible. But in your case, we have $0^\#$, from which we are able to define a truth predicate for firstorder truth in $L$, and so your question can be formulated. But see the comments at the bottom concerning an ambiguity about this in nonstandard models.
For each natural number $n$, the least $\Sigma_n$correct cardinal $\kappa_n$ is definable in $L$, meaning $L_{\kappa_n}\prec_{\Sigma_n}L$, since we can express this property using a $\Sigma_n$ truth predicate. Further, any $\Sigma_n$ definable ordinal will be bounded by $\kappa_n$. Thus, $\kappa=\sup_n\kappa_n$.
The union of an increasingly elementary chain is elementary, so $L_\kappa\prec L$. Thus, the supremum of the definable ordinals is the same as the first fully correct cardinal.
But this is never a Silver indiscernible, since (as Monroe points out in the comments below) Silver indiscernibles $\xi$ are inaccessible in $L$ and therefore have many smaller $L_\alpha\prec L_\xi$. So the least Silver indiscernible is strictly larger than the least fully correct cardinal in $L$.
This answers your question, but let me augment my answer with the following observation for the general ZFC case, which I find interesting.
Theorem. For any model $M$ of ZFC, not necessarily wellfounded, let $W$ be the collection of all $x\in M$ for which $x\in (V_\alpha)^M$ where $\alpha$ is an ordinal definable in $M$ without parameters. This is called the definable cut of $M$. Then $W\prec M$.
Proof. Notice that the definable cut may not necessarily have a least upper bound in $M$. It could be that $W=M$ or that the supremum of $W$ is not realized in $M$.
But we can nevertheless prove that $W$ is an elementary substructure of $M$ by verifying the TarskiVaught criterion. If $M$ has a witness for an extensional statement with parameters in $W$, then the least rank of such a witness is definable from those parameters, and by considering all possible parameters up to a definable rank, we can get a definable bound on the rank of the witnesses. And so the witness is in $W$. So we have fulfilled the TarskiVaught criterion, and thus $W\prec M$. $\quad\Box$
Note that in this theorem, we are using the external notion of definability, coming from outside the model, whereas in your question, we were using the internal notion of definability provided by the truth predicate defined from $0^\#$. These are not necessarily the same, even when $0^\#$ exists, since there could be $\omega$nonstandard models of ZFC with $0^\#$. This issue can be seen as a possible ambiguity in your question, as to whether you intend to use the internal notion of definability or the metatheoretic notion of definability.
Observation. If ZFC+$\exists 0^\#$ is consistent, then there is a model of this theory in which the supremum of the definable$L$withoutparameters ordinals (understood using metatheoretic definitions) does not exist.
Proof. Take any $\omega$nonstandard model of the theory. For standard $n$, the first $\Sigma_n$correct cardinal of $L$ is definable in $L$, using a standardfinite definition. The supremum of these would be the definable cut of the model, and this is the supremum of the metatheoretically definable ordinals in $L$. This is bounded in the ordinals of $L$, because by overspill there must be some $\Sigma_n$ correct ordinals for nonstandard $n$. The definable cut can have no supremum in the model, since from it we could define the standard cut in $\omega$, which is possible. $\quad\Box$

2$\begingroup$ If $\xi$ is a Silver indisernible, then it is inaccessible in $L$, so there are many $\alpha<\xi$ such that $L_\alpha \prec L_\xi$. Thus $L_\kappa \prec L$ does not imply $\kappa$ is an indiscernible. $\endgroup$ Jun 3 at 9:59

$\begingroup$ Of course we should talk in the standard model, because in a nonstandard model the supremum may not exist. $\endgroup$ Jun 3 at 10:26

1$\begingroup$ Because then $\alpha$ would also be fully correct, but $\kappa$ was the first fully correct ordinal. $\endgroup$ Jun 3 at 11:14

1$\begingroup$ Thank you, now I've understood it. Then, how could we understand how large the least Silver indiscernible is? For example, are there some "canonical" sequence of ordinal of length $\omega$ with limit the least Silver indiscernible? $\endgroup$ Jun 3 at 11:19

1$\begingroup$ @Reflecting_Ordinal $i_0$ is the sup of countable ordinals which are definable in $L$ from $V$cardinals (equivalently, from $\omega_n^V$'s). $\endgroup$ Jun 3 at 16:56