Working in with the constructible hierarchy, let's define $Birth$ and $Rank$ of sets as:
$Birth(x)=\alpha \iff min \ \alpha: x \in L_{\alpha+1}$
$Rank(x)=\alpha \iff min \ \alpha : x \subseteq L_\alpha $
Let's call sets $appropriate$, if and only if their birth is their rank, for example $\omega$.
Now if we build a new hierarchy $L'$ that requires sets to be not just constructible from below, but also appropriate. Lets denote the stages of that hierarchy by $L'_\alpha$, so each successor stage $L'_{\alpha+1}$ is the set of all appropriate constructible subsets of the prior stage $L'_\alpha$. Of course $L'$ is the union of all appropriate constructible successor and limit stages.
Question 1: Is $L'$ is an inner model of ZFC?
Question 2: is $L'_\omega = L_\omega $?
Best Answer
You cannot possibly expect to have $L'$ as a model of $\sf ZF$, since it is clearly smaller than $L$.
To see why, note that your requirement shows that $\mathcal P(\omega)^{L'}\subseteq L_{\omega+1}$. But this set is countable in $L$, and so certainly not the power set of $\omega$ itself.