= Solution
{c}
Induct on $\alpha$ using the transitivity and inclusion already established. The zero stage is immediate. Suppose $L_\alpha\cap\operatorname{Ord}=\alpha$. Being an <ordinal> is expressible by a <bounded formula in set theory>, for example transitivity, transitivity of all members, and membership comparability. It is therefore absolute for the transitive set $L_\alpha$. The subset of its elements that are ordinals is internally definable, so
$$
\alpha=\{x\in L_\alpha:x\text{ is an ordinal}\}\in\operatorname{Def}(L_\alpha,1)=L_{\alpha+1}.
$$
By inclusion, all smaller ordinals are present as well. Conversely, if an ordinal $\gamma$ belongs to $L_{\alpha+1}$, then $\gamma\subseteq L_\alpha$, since every member of a <definable power set> is a subset of the preceding domain. All members of $\gamma$ are ordinals, so $\gamma\subseteq L_\alpha\cap\operatorname{Ord}=\alpha$ and $\gamma\leq\alpha$. Hence the ordinal part of the successor level is exactly $\alpha+1$.
At a nonzero limit ordinal $\lambda$,
$$
L_\lambda\cap\operatorname{Ord}=\bigcup_{\alpha<\lambda}(L_\alpha\cap\operatorname{Ord})=\bigcup_{\alpha<\lambda}\alpha=\lambda.
$$
Thus
$$
\boxed{L_\alpha\cap\operatorname{Ord}=\alpha\quad\text{for every ordinal }\alpha.}
$$
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