ans is A
$First(S)=First(C)=\{c,d\}$
There are no multiple entries in single row of parsing table hence grammar is LL1
Note : If we have $A \rightarrow B\mid C,$ for grammar to be LL(1) first(B) intersection First(C) should be null otherwise grammar is not LL1. If First(B) contains $\epsilon$ then Follow(A) intersection First(C) should be null. Using this we can say grammar is LL(1) or not without constructing parsing table.
An $\epsilon$ free LL(1) grammar is also SLR(1) and hence LALR(1) and LR(1) too.