Correct Answer - C
First two gate are `NOT` gates and the last gate is `NOR` gate
Thus, `Y_(1)=bar(A),y_(2)=bar(B)` and `y=bar(y_(1)+y_(2))`
The truth table corresponding to this is as follows:
`|(A,B,y_(1)=vec(A),y_(2)=vecB,y_(1)+y_(2),y=bar(y_(1)+y_(2)),A.B),(0,0,1,1,1,0,0),(0,1,1,0,1,0,0),(1,0,0,1,1,0,0),(1,1,0,0,0,1,1)|`
Thus the combination of gate represents `AND` gate.