We can get a DFA for $L = \{x \mid xx ∊ A\}$ as follows:
Take DFA for $A$ $\left(Q, \delta, \Sigma, S, F\right)$ with everything same except initially making $F = \phi$.
Now for each state $D \in Q$, consider 2 separate DFAs, one with $S$ as the start state and $D$ as the final state and another with $D$ as the start state and set of final states $⊆ F$. If both these DFAs accept same language make $D$ as final state.
This procedure works as checking the equivalence of 2 DFAs is decidable.
Contradictions for other choices
a) Consider $A = Σ^* $. Now for $ w \in A, L = \{xx \mid x \in A\} = \{ww \mid w \in Σ^*\} $ which is context sensitive
c) Same example as for (a)
d) Consider $A = \{a^nb^n c^* a^*b^mc^m \mid n, m \ge 0\} $
This is CFL. But if we make $L$ from $A$ as per (d), it'll be
$L = \{a^nb^nc^n \mid n \ge 0\}$ which is not context free..