option a is answer
we can make only one parse tree for every string generated by this grammar , moreover the language generated by this grammar is set of all the even length palindrome
so it is unambiguous
now it is not LL(1)
because S ⇒ aSa | bSb |ε
here the follow of S is a,b ,$ that is match with the first of right hand side
it means is row S and column a contain 2 entry and same thing will happen for column b watch the diagram
|
a |
b |
$ |
|
S |
S->aSa,S->epsilon |
S->bSb,S->epsilon |
S->epsilon |
|
now the given grammar is not DCFL so it is not LR(k)