Let us accept the von Neumann cardinal assignment for this question. Furthermore, given a cardinal number $\kappa$, let us write $2^\kappa$ for the unique cardinal number isomorphic to the powerset of $\kappa$ (so in particular, we're doing cardinal exponentiation, not ordinal).
Also, write $C$ for the subclass of the ordinals consisting of all cardinal numbers. Then we have the following observation. $$\forall \kappa \in C :\mathrm{ord}(\kappa \cap C) \leq \kappa.$$
Question. Does ZFC prove the following? For all $\kappa \in C$, the following are equivalent.
- $\mathrm{ord}(\kappa \cap C) = \kappa$
- $\kappa$ is weakly inaccessible
I think I have the backward direction. (Is it correct?)
$(\leftarrow)$ Proof. Suppose $\kappa$ is a weakly inaccessible cardinal number. Then since $\kappa$ is a limit cardinal, hence $\kappa \cap C$ is cofinal in $\kappa$. Hence $|\kappa \cap C| = \kappa,$ since $\kappa$ is regular. Therefore $\mathrm{ord}(\kappa \cap C) \geq \kappa.$ Hence by the observation preceding the question, it follows that $\mathrm{ord}(\kappa \cap C) = \kappa.$
No, there are singular cardinals $\kappa$ such that $\kappa\cap C$ has order-type $\kappa$. The first such $\kappa$ is the supremum of the sequence (of length just $\omega$) defined inductively by $\xi(0)=\aleph_0$ and $\xi(n+1)=\aleph_{\xi(n)}$.