Given, $L = \{ab, aa, baa\}$
$L^0 = \{\epsilon\}$
$L^1 = \{ab, aa, baa\}$
$L^2 = L^1.L^1$ $=$ $\{abab, abaa, abbaa, aaab,aaaa, aabaaa, baaab, baaaa, baabaa\}$
..
$L^n$ = $3^n$ possible strings by arranging any of the 3 strings in $L^1$.
As we know,
$L^* = L^0$ $\cup$ $L^1$ $\cup$ $L^2$ $\cup$ $L^3$ $\cup$ $….$
Now, coming to options,
(B) abaabaaabaa = ab+aa+baa+ab+aa
(C) aaaabaaaa = aa+aa+baa+aa
(D) baaaabaa = ba+aa+ab+aa
Only, option (A) cannot be created using $L^*$.