Here by elimination technique we can work out.Let us see how..
We consider the minimal string which is a according to the grammar due to the production S --> a and hence epsilon is not produced.However , both option A) and B) options' regular expressions accepts epsilon which is wrong accoding to the given grammar.Hence options A) and B) can be eliminated straightaway.
Now coming to option C) , we know minimal string generated by the grammar is 'a' now using the production S --> SSS we can have aaa as new string which satisifies the given regular expression.Similarly substituting one S as 'a' , one S as 'aaa' and final S as 'ab' , we get the new string as 'aaaaab' which is also satisfied by the given regular expression.
Similarly proceeding we can generate the strings from the grammar and then validate using the given regular expression.If even a single violation is found , we say the regular expression for the given grammar is incorrect.
But here no violation occurs.
Hence C) should be the correct option.