This is a follow-up to my questions here and here. Let $X= (A,+,*,<)$ be an ordered field. Let us define a constructible hierarchy relative to $X$ as follows. Let $D_0(X)=A\cup A^2 \cup \{+,*,<\}$. For any ordinal $\beta$, let $D_{\beta+1}(X)=Def(D_{\beta+1}(X))$. For any limit ordinal $\gamma$, let $D_\gamma(X)=\cup_{\beta<\gamma}D_\beta$. Finally, let $D(X)=\cup_\alpha D_\alpha$.
Suppose that $D(X)=V$, and that $X$ is not Dedekind-complete. Then my question is, what is the least $\alpha$ such that $D_\alpha(X)$ sees that $X$ is not Dedekind-complete, that is, such that $D_\alpha(X)$ contains a nonempty bounded-above subset of $X$ with no least upper bound in $X$?