G:S$\rightarrow$aA
A $\rightarrow$ BB
B$\rightarrow$ aBb /∈
(A) Here S can generate a (s->aA->aBB->aB->a) which can not generate by A.So we can eliminate this option.
(B)In this option we can not able to generate aab which can be generate by given grammar.So we can eliminate this option.
(C)In this option we can generate a. we can't generate anything other than a because there is no terminal present to stop B.So we can eliminate this option.
Answer should be (D)G:S$\rightarrow$aA , A $\rightarrow$ BB/B , B$\rightarrow$ aBb /ab.