You are asked to express the sets in terms of union, intersection, complements.
In particular, the "minus" in your proposed solution would have to be expressed in terms of these. But aside from that minor comment, your answer is correct.
We do the problem in another way. We look at the four shaded parts, one after the other, going roughly downwards.
The top shaded part is inside $A$, inside $B$, outside $C$, outside $D$, and outside $E$.
In symbols, it is
$$A\cap B\cap C^c\cap D^c\cap E^c,$$
($S^c$ denotes the complement of $S$).
The next one down is very similar, except we are outside $B$, inside $C$, with the rest exactly as above:
$$A\cap B^c\cap C\cap D^c\cap E^c.$$
The next one down is outside $A$, outside $B$, and inside $C$, $D$, and $E$:
$$A^c\cap B^c\cap C\cap D\cap E.$$
Finally, the one at the bottom is inside $E$, and outside all the others:
$$A^c\cap B^c\cap C^c\cap D^c\cap E.$$
Finally, we take the union of our four pieces. It is long, so we use a smaller font:
$$\small(A\cap B\cap C^c\cap D^c\cap E^c)\cup(A\cap B^c\cap C\cap D^c\cap E^c)\cup(A^c\cap B^c\cap C\cap D\cap E)\cup(A^c\cap B^c\cap C^c\cap D^c\cap E).$$
Comment: This was a bit lengthy, but utterly mechanical. There are many other forms for the answer. For the purpose of circuit design, we would almost certainly want an expression that used fewer symbols. (We used a total of $19$ $\cap$ and/or $\cup$. The complement operations were not counted, because in circuit design they are often assumed to be "for free.")