Solution

Gate $1$ is a NAND gate with inputs $A$ and $B$.

So, output of gate $1$ is

$\overline{AB}$

Now, gate $2$ is a NOR gate.

Its inputs are $\overline{A}$ and $B$.

So, output of gate $2$ is

$\overline{\overline{A}+B}$

Using De Morgan's law,

$\overline{\overline{A}+B}=A\overline{B}$

Now, gate $3$ is also a NOR gate.

Its inputs are $\overline{AB}$ and $A\overline{B}$.

So,

$Y=\overline{\overline{AB}+A\overline{B}}$

Now,

$\overline{AB}=\overline{A}+\overline{B}$

Therefore,

$Y=\overline{\overline{A}+\overline{B}+A\overline{B}}$

Since $\overline{B}+A\overline{B}=\overline{B}$,

$Y=\overline{\overline{A}+\overline{B}}$

Using De Morgan's law,

$Y=AB$