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Solution

$L=\{ab,aa,baa\}$

Check option (b):

$abaabaaabaa=ab\cdot aa\cdot baa\cdot ab\cdot aa$

So, option (b) is in $L^*$.

Check option (c):

$aaaabaaaa=aa\cdot aa\cdot baa\cdot aa$

So, option (c) is in $L^*$.

Check option (d):

$baaaaabaa=baa\cdot aa\cdot ab\cdot aa$

So, option (d) is in $L^*$.

But option (a) cannot be completely divided into strings $ab,aa,baa$.

Therefore, $baaaaabaaaaa$ is NOT in $L^*$.