The new question added a few minutes ago can be answered by the same idea as in Emil's comment. The following paragraph is provable in ZFC and therefore true in HOD:
For any cardinal $\kappa\leq\mathfrak c$ (the cardinal of the continuum), there is a real-closed subfield of $\mathbb R$ with transcendence degree $\kappa$, and by adjoining $i$ to this field we get an algebraically closed field of transcendence degree $\kappa$. Since all algebraically closed fields of characteristic 0 and transcendence degree $\kappa$ are isomorphic, they all have real-closed Archimedean subfields of index 2. On the other hand, an algebraically closed field of transcendence degree $\kappa>\mathfrak c$ cannot have such a subfield, because the subfield would have cardinality $>\mathfrak c$ and would therefore admit no embedding in the reals, so it can't be archimedean.
Using this information in HOD and the fact that archimedeanness is absolute between HOD and the universe (essentially because the natural numbers are absolute), we get that an HOD algebraically closed field of characteristic 0 has an archimedean real-closed HOD subfield of index 2 if and only if its transcendence degree in HOD (or equivalently its cardinality in HOD) is at most the cardinal that HOD considers to be the cardinality of the continuum.