The truth table for the given circuit is

From the given logic circuit, the boolean expressions for the outputs of the two AND gates are:
$C = A \cdot B$
$D = \bar{A} \cdot B$

The final output $E$ from the OR gate is:
$E = C + D$
$E = A \cdot B + \bar{A} \cdot B$

Taking $B$ common, we get:
$E = B(A + \bar{A})$
Since $A + \bar{A} = 1$,
$E = B \cdot 1$
$E = B$

Therefore, the output $E$ is identical to the input $B$.
The corresponding truth table is:
$A = 0, B = 0 \Rightarrow E = 0$
$A = 0, B = 1 \Rightarrow E = 1$
$A = 1, B = 0 \Rightarrow E = 0$
$A = 1, B = 1 \Rightarrow E = 1$

Answer:

A	B	E
0	0	0
0	1	1
1	0	0
1	1	1