A. $A \to C$, $C$ is non-key attribute. $AB$ is candidate key and so is a partial FD. Hence, not in 2NF.
B. Here, $C\to B$ makes $AC$ also a candidate key. So, $B \to C$ and $C \to B$ are not violating 2NF as they are detrmining "key" attribute. But $A \to D$ violate 2NF.
C. $AB$ and $DB$ are candidate keys here. Due to $C \to D$, $CB$ also becomes a candidate key. So, no other attribute present other than prime(key) attributes. So, $R$ is in 2NF as well as 3NF.
D. Only $AB$ is the candidate key. $BC$ is not a super key, is also not a proper subset of candidate key. $D$ is not a prime attribute. So, $BC \to D$ violate 3NF but due to no partial dependency, $R$ is in 2NF.
So, D is the answer.