Show that this language is decidable?
If $E_{DFA}$ with input $C$ accepts, then $C = L(M) \cap B = \emptyset$.
In other words, no word containing an odd number of 1s is also in $L(M)$.
Then, by definition of $A$ you should accept.
Conversely, if  $E_{DFA}$ with input $C$ rejects, then $C = L(M) \cap B  \neq \emptyset$.
That is, there is some word $w \in L(M)$ such that $w$ contains an odd number of 1s. Then, by definition of $A$, you should reject.