Taking the 1st IP Address: 128.96.171.92
Bitwise AND between $128.96.171.92$ and $255.255.254.0$ we get the subnet ID as follows:$$ \begin{array} {rrrr}255&255&11111110&0 \\ 128&96&10101011&92 \\ \hline \textbf{128}&\textbf{96}&\textbf{10101010}&\textbf{0}\end{array}$$
∴ Subnet ID = 128.96.170.0
∴ 128.96.171.92 will forward to interface 0
Taking the 2nd IP Address: 128.96.167.151
Bitwise AND between $128.96.167.151$ and $255.255.254.0$ we get, $$ \begin{array} {rrrr}255&255&11111110&0 \\ 128&96&10100111&151 \\ \hline \textbf{128}&\textbf{96}&\textbf{10100110}&\textbf{0}\end{array}$$
∴ Subnet ID = 128.96.166.0
∴ 128.96.167.151 will forward to interface R2
Taking the 3rd IP Address: 128.96.163.151
Bitwise AND between $128.96.163.151$ and $255.255.254.0$ we get, $$ \begin{array} {rrrr}255&255&11111110&0 \\ 128&96&10100011&151 \\ \hline \textbf{128}&\textbf{96}&\textbf{10100010}&\textbf{0}\end{array}$$
∴ Subnet ID = 128.96.162.0 (Doesn’t match with any given interface)
Now, Bitwise AND between $128.96.167.151$ and $255.255.252.0$ we get, $$ \begin{array} {rrrr}255&255&11111100&0 \\ 128&96&10100011&151 \\ \hline \textbf{128}&\textbf{96}&\textbf{10100000}&\textbf{0}\end{array}$$
∴ Subnet ID = 128.96.160.0 (Doesn’t match with any given interface)
∴ 128.96.163.151 will forward to default interface R4
Taking the last IP Address: 128.96.164.121
Bitwise AND between $128.96.164.121$ and $255.255.254.0$ we get, $$ \begin{array} {rrrr}255&255&11111110&0 \\ 128&96&10100100&121 \\ \hline \textbf{128}&\textbf{96}&\textbf{10100100}&\textbf{0}\end{array}$$
∴ Subnet ID = 128.96.164.0
∴ 128.96.167.151 will forward to interface R3
∴ Option (A) is the correct answer.