To get the truth table shown, from the following logic circuit, the Gate \(\mathrm{G}\) should be

\(\begin{array}{|c|c|c|}\hline\mathrm{A} & \mathrm{B} & \mathrm{Y} 0 & 0 & 0 0 & 1 & 0 1 & 0 & 1 1 & 1 & 1\\\hline\end{array}\)

Truth table for \(\mathrm{Y}\), with the possible values of \(\mathrm{C}\) is,
\(\begin{array}{|c|c|c|}
\hline \mathbf{A} & \mathbf{C} & \mathbf{Y} \\
\hline 0 & 0 & 0 \\
\hline 0 & 0 & 0 \\
\hline 1 & 0,1 & 1 \\
\hline 1 & 0,1 & 1 \\
\hline
\end{array}\)
For gate \(\mathrm{G}\)
\(\begin{array}{|c|c|c|c|}
\hline & A & B & C \\
\hline (I) & 0 & 0 & 0 \\
\hline (II) & 0 & 1 & 0 \\
\hline (III) & 1 & 0 & 0,1 \\
\hline (IV) & 1 & 1 & 0,1 \\
\hline
\end{array}\)
\(\mathrm{G}\) is not a NOT gate as NOT gate takes only one input. (II) indicates \(\mathrm{G}\) is not a OR gate as OR gate would give high output for the inputs in (II). Also, (II) indicates it is not a XOR gate as XOR would also give high output for inputs in (II). Hence, the given truth table is satisfied only by AND gate.