The circuit consists of two NOT gates followed by a NOR gate. Let the inputs be A and B. The outputs of the NOT gates are $\bar{A}$ and $\bar{B}$. These are inputs to a NOR gate, so $Y = \overline{\bar{A} + \bar{B}}$. By De Morgan's law, $Y = \overline{\bar{A}} \cdot \overline{\bar{B}} = A \cdot B$. This is an AND gate. The truth table for an AND gate is (0,0)->0, (0,1)->0, (1,0)->0, (1,1)->1. This matches option (2).