Claim. The following are equivalent for a limit ordinal $\alpha$ :
$\alpha$ is a beth-fixed point.
$V_\alpha$ thinks every set is equipotent with an ordinal.
For $1\to 2$, it suffices to show that for every $\beta<\alpha$, $V_\alpha$ thinks $V_\beta$ is equipotent with an ordinal. (This is because every set in $V_\alpha$ is a subset of some $V_\beta$, $\beta<\alpha$.) Let $f:V_\beta\to\beth_\beta$ be a bijection. Then $f$ is a subset of $V_\beta\times \beth_\beta$, which is a member of $V_\alpha$. Since $V_\alpha$ is closed under Cartesian products and power sets, $f\in V_\alpha$.
For $2\to 1$, observe that the assumption implies $|V_\beta|\in V_\alpha$ for every $\beta<\alpha$, so $\beth_\beta<\alpha$ for all $\beta<\alpha$, which means $\alpha$ is a $\beth$-fixed point.
I finally prove that the above characterization is equivalent to the validity of $\Sigma_1$-replacement over $V_\alpha$:
Claim. If $\alpha$ is a beth-fixed point, then $V_\alpha$ satisfies $\Sigma_1$-Replacement.
The main ingredient is the following version of Levy reflection principle (which is provable by the same proof of usual Levy reflection principle $H_\kappa\prec_{\Sigma_1} V$)
Theorem. Let $\lambda<\kappa$ be cardinals and $\lambda$ be regular. Then $H_\lambda\prec_{\Sigma_1} V_\kappa$.
Moreover, it is known that $H_\lambda$ is a model of ZFC without Power set if $\lambda$ is regular.
Now let $F$ be a $\Sigma_1$-class function over $V_\alpha$ with a parameter $p$. Take $x\in V_\alpha$. Choose $\xi<\alpha$ such that $p,x\in V_\xi$. Since $\alpha$ is a beth-fixed point, $\lambda:=|V_\xi|^+<\alpha$. We can see that $V_\xi\subseteq H_\lambda\subseteq V_\alpha$.
Observe that $F$ is absolute between $V_\alpha$ and $H_\lambda$. Moreover, $H_\lambda$ satisfies Replacement for $F$. Let $H_\lambda\models F^"[x]=y$ for $y\in H_\lambda$. Since the formula
$$[\forall v\in y\exists u\in x (F(u)=v)]\land [\forall u\in x\exists v\in y (F(u)=v)]$$
is $\Sigma_1$-formula, it also holds over $V_\alpha$. This shows $y$ witnesses the instance of replacement for $F$, $x$ and $p$.
The answer is negative. I claim that the supremum can see the
$\omega$-cofinal sequence $\kappa_n$, where $\kappa_n$ is the least
$\Sigma_n$-extendible cardinal.
The main point is that if $\alpha$ is $\Sigma_{n+1}$-extendible, so
$V_\alpha\prec_{\Sigma_{n+1}}V_\beta$ for some $\beta$, then
$V_\beta$ can see that $\alpha$ is $\Sigma_n$-extendible, and so
the least $\Sigma_n$-extendible is in fact less than $\alpha$. So
each next $\Sigma_{n+1}$-extendible cardinal must be beyond the $\beta$ used for
the $\Sigma_n$-extendibility of the least $\Sigma_n$-extendible cardinal. Therefore, the limit $\kappa$ of the
least $\Sigma_n$-extendibles can see that they are like that, so by restricting the definition of $\prec_{\Sigma_{n+1}}$ just to $V_\beta$
it can define the cofinal $\omega$-sequence, which means it
isn't even $\Sigma_2$-extendible.
Another argument: if $\kappa$ is the least worldly cardinal, then let $\delta<\kappa$ be $\Sigma_2$-correct in $V_\kappa$, which is possible, because ZFC proves that the $\Sigma_2$-correct cardinals form a club. But $V_\delta$ is correct about whether a cardinal is $\Sigma_n$-extendible, and whether there is one. So the whole sequence is below $\delta$.
Best Answer
If $V_\alpha$ is a model of ZFC, then $\alpha$ must be a cardinal, and much more. In fact it must be a strong limit cardinal, a $\beth$-fixed point, a fixed point in the enumeration of $\beth$-fixed points, and have any other strong limit property of this sort.
To see this, observe that ZFC proves that the $\beth$-hierarchy is unbounded, but also we can show that the $\beth$ hierarchy (and the Von-Neumann hierarchy) is absolute for a "full" model of the form $V_\alpha,$ since the model's power set operator is the same as the real one. So it follows that $\alpha$ is a strong limit. And the stronger properties follow from similar considerations.
In a little more detail, if $\beta <\alpha,$ then $P(\beta) \in V_\alpha$ since $P(\beta)$ is just two ranks higher than $\beta,$ and so since being a subset is absolute, $P(\beta)^{V_\alpha}=P(\beta).$ ZFC proves $2^{|\beta|}$ exists, which relativized to $V_\alpha$ is the least ordinal in $V_\alpha$ that has a bijection in $V_\alpha$ with $P(\beta).$ And being a bijection is absolute so this is a real bijection, thus there is an ordinal in $V_\alpha$ that is in one-to-one correspondence with $P(\beta),$ so $2^{|\beta|}<\alpha.$