Actually the answer given is wrong.
Neither of them is correct.
(a+b)* also contains those string which have a after b or have a substring of abab and many more permutations.
The correct ans would be {anbm | m,n>=0}
Proof:
L1: anbm | n,m >=0
L2 : anbm| m=n
L2' : anbm | m!=n
Now L2' is a subset of L1 as L1 will contain all the string of form a^nb^n,
SetA (union) subset of SetA = SetA
Therefore L will be L1 only.