The input power $P_{\text{in}}$ is the sum of the power developed $P_{\text{dev}}$ and the copper losses $P_{\text{cu}}$, plus the constant iron and friction losses $P_{\text{cf}}$. Given $V_{\text{L}} = 400 \text{ V}$, $I_{\text{a}} = 150 \text{ A}$, $R_{\text{a}} = 0.4 \text{ } \Omega$, $P_{\text{dev}} = 130 \text{ kW}$, and $P_{\text{cf}} = 1 \text{ kW}$, we calculate copper loss as $3 \times I_{\text{a}}^2 \times R_{\text{a}} = 3 \times 150^2 \times 0.4 = 27 \text{ kW}$. The total output power is $P_{\text{out}} = P_{\text{dev}} - P_{\text{cf}} = 130 - 1 = 129 \text{ kW}$. The input power is $P_{\text{in}} = P_{\text{dev}} + P_{\text{cu}} = 130 + 27 = 157 \text{ kW}$. The efficiency $\eta$ is $\frac{P_{\text{out}}}{P_{\text{in}}} \times 100\%$.

Always ensure you distinguish between power developed (internal) and output power (shaft) when dealing with efficiency problems in rotating machines.