Consider a system of states with a binary relation R. If R(x,y) holds, we say that state x reaches state y. Further, consider two unary predicates I and B where I(x) means that x is an initial state and B(x) means that x is a bad state.
Question: States that reach only bad states are bad.
Answer: ∀x(∀y(R(x,y) → B(y) → B(x)))
However, my question is in regard to "-> B(y)" shouldn't that be an and (∧),
∀x(∀y(R(x,y) ∧ B(y) → B(x)))
First, you need to add some parentheses, because $P \rightarrow (Q \rightarrow R)$ is not the same as $(P \rightarrow Q) \rightarrow R)$
Indeed, I see that you are hoping to use the Exportation equivalence rule, which states that
$$P \rightarrow (Q \rightarrow R) \Leftrightarrow (P \land Q) \rightarrow R$$
so that means that your intended symbolization must have been:
$$\forall x (\forall y (R(x,y) \rightarrow (B(y) \rightarrow B(x))) \tag{*}$$
Unfortunately, $(*)$ is not the correct symbolization of the original sentence. $(*)$ translates back as: For any two states that you have, if the first reaches the second, then if the second is bad, then the first is bad. So, that would mean that as soon as the first state reaches any bad state, then the first state is bad. That is not the same as the original sentence, since that is only talking about states that only reach bad states. So, for example, take a state that reaches both good and bad states. Then by $(*)$, it must be a bad state, since it reaches at least one bad state. But the original English sentence has nothing to say about states that each both good and bad states.
OK, so how do you translate the original English sentence? Well, follow this form:
$$\forall x (\text{"x reaches only bad states"} \rightarrow B(x))$$
OK, and the translation of "x reaches only bad states" is:
$$\forall y (R(x,y) \rightarrow B(y)$$
So, plug this in, and you get:
$$\forall x (\forall y (R(x,y) \rightarrow B(y)) \rightarrow B(x))$$