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The given grammar is LL(3).

Some prorperties of LL grammar are that it should not be ambiguous, non determinstic and left recursive.

So if we see for LL(1): it has a look ahead of 1 so that makes this grammar non deterministic as after seeing a we dont whether it is S->a or S->ab or S->abc.. So not LL(1)

LL(2): after seeing ab we dont know S->ab or S->abc.

LL(3) has a lookahead of 3 and this problem doesnot reside here. So the given grammar is LL(3).

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I think that the given grammar is LL(1) . please explain me if I'm wrong.