In given, the logic relation
$$
\mathrm{Y}=\overline{\mathrm{A}} \cdot \mathrm{B}+\mathrm{A} \cdot \overline{\mathrm{B}}=\mathrm{Y}_1+\mathrm{Y}_2
$$
So, $Y_1=A \cdot B$ and $Y_2=A$. $\bar{B}$
$Y_1$ can be obtained as output of AND gate I for which one Input is A through NOT gate and another input is of B. $Y_2$ can be obtained as output of AND gate II for which one input is of $A$ and other input is of $B$ through NOT gate.
So, $\quad Y_1=\bar{A} \cdot B, Y_2=A . \bar{B}$
So, the given table can be obtianed form the logic circuit given as :-


Now, $Y_1$ and $Y_2$ are feed into the two terminals of $O R$ gate to get $(\mathrm{Y})$.
$$
\mathrm{Y}=\mathrm{Y}_1+\mathrm{Y}_2 \text { or } \mathrm{Y}=\overline{\mathrm{A}} \cdot \mathrm{B}+\mathrm{A} \cdot \overline{\mathrm{B}}
$$