$Y_{1}=A+B ; Y_{2}=A+B$, as gate 1 and 2 are OR gates.
As gate 3 is NAND gate, its output will be
$Y =\overline{Y_{1} Y_{2}}=\bar{Y}_{1}+\bar{Y}_{2}=\overline{A+B \cdot A+B}$
$=\overline{A+B}+\overline{A+B}=\overline{A+B}$
Using De-Morgan's theorem and identity $A+A=A$
So, given output is that of NOR gate.
