To get the truth table shown from the following logic circuit, the gate G should be

From the given logic circuit, the output $Y$ is given by the Boolean expression:

$Y = A + G(A, B)$

From the given truth table, we can observe that the output $Y$ is identical to the input $A$ for all combinations of $A$ and $B$.

Thus, $Y = A$.

Equating the two expressions for $Y$, we get:

$A + G(A, B) = A$

Let us check the given options for gate G:

If G is an AND gate, $G(A, B) = A \cdot B$.

Then, $Y = A + A \cdot B = A(1 + B) = A \cdot 1 = A$.

This perfectly matches the required condition.

Checking other options:
If G is an OR gate, $Y = A + (A + B) = A + B \neq A$.
If G is a NAND gate, $Y = A + \overline{A \cdot B}$. For $A=0, B=0$, $Y = 0 + 1 = 1 \neq 0$.
If G is a NOR gate, $Y = A + \overline{A + B}$. For $A=0, B=0$, $Y = 0 + 1 = 1 \neq 0$.

Therefore, the gate G must be an AND gate.

Answer: AND gate.