Answer : 2 (2NF)
Given Relation schema $R(ABCD)$ and $FD$ set $\left \{ AB \rightarrow CD, BC \rightarrow D \right \}$
We can find the Candidate key(s) of the Relation $R$ using the FD Set and the Candidate key of the relation $R$ will be $AB$ .
Hint : If some Attribute(s) are Not on the Right Side of any Non-trivial FD, then those attributes must be the Part of Every Candidate key.
Checking for $2NF$ :
Part of the Candidate key $AB$ i.e. $A$ or $B$ are Not deriving any Non-prime attribute, Hence, No Partial Dependency. Hence, the Relation $R$ is in $2NF$.
Checking for $3NF$ :
Look at the Non-trivial FD $BC \rightarrow D$.. It is a Transitive dependency as the combination of part of the candidate key($B$) along with a non-prime attribute($C$) is deriving a Non-key attribute $D$. Hence, We can say that Non-prime attribute $D$ is Transitively dependent on candidate key $AB$. So, there is Transitive Dependency. Hence, the Relation $R$ is Not in $3NF$.
Refer here for $2NF$ and $3NF$ : https://gateoverflow.in/219972/normalization?show=220006#a220006