Consider the circuit in figure. $f$ implements
$\overline{A} \overline{B}C + \overline{A}B \overline{C} + ABC$
$A + B + C$
$A \oplus B \oplus C$
$AB + BC + CA$
So, $f = \bar A \bar B C + \bar A B \bar C +A \bar B \bar C + ABC$ $\qquad = \bar A (\bar BC + B \bar C) + A (\bar B \bar C + BC)$ $\qquad = \bar A (B \oplus C) + A (B \odot C)$ $\qquad = \bar A (B \oplus C) + A (\overline{B \oplus C})$ $\qquad = A \oplus B \oplus C$
Correct Answer: $C$
See the last part...