Given initial speed S1 = 40 km/hr and initial time T1 = 3 hrs.

The formula for distance is $D = S \times T$, where $D$ is distance, $S$ is speed, and $T$ is time.
Given initial speed $S_1 = 40 \text{ km/hr}$ and initial time $T_1 = 3 \text{ hrs}$.

$$D = 40 \text{ km/hr} \times 3 \text{ hrs}$$ $$D = 120 \text{ km}$$

We need to find the new speed $S_2$ to cover the distance $D = 120 \text{ km}$ in new time $T_2 = 2 \text{ hrs}$.

$$S_2 = \frac{D}{T_2}$$ $$S_2 = \frac{120 \text{ km}}{2 \text{ hrs}}$$ $$S_2 = 60 \text{ km/hr}$$

The train must travel at an average speed of:

$$\mathbf{60 \text{ km/hr}}$$