First of all LTL is defined over traces. In verification, we typically say that a state satisfies some LTL formula if all of the traces starting in the state satisfy the formula. I henceforth assume that this is what is meant in the question and this definition has been given in your course.
The Until operator is defined over a word as follows:
$$
w \models \psi \mathcal{U} \phi \quad \text{ iff } \quad \exists i \in \mathbb{N}: \forall j\leq i. w^i \models \psi \wedge m^j \models \phi
$$
In this equation, $w^i$ represents the word starting at position $i$, where all letters before position $i$ have been removed. Applied to the concretization $\psi = b$ and $\phi = b$, we have:
$$
w \models b \mathcal{U} b \quad \text{ iff } \quad \exists i \in \mathbb{N}: \forall j\leq i. w^i \models n \wedge m^j \models b
$$
Note that without loss of generality, we can assume that the $i$ searched for in the definition is always 0, because for $j=0$, we need to have $w^0 \models b$ for a word $w$ to be a model of $b \mathcal{U} b$. But then we can also choose $i=0$. Then the definition simplifies to:
$$
w \models b \mathcal{U} b \quad \text{ iff } \quad m^0 \models b
$$
Hence, you are searching for all states for which all outgoing paths start with a $b$ state, which is equivalent to saying that you search for all states in which $b$ hold.
