A train travels the first one-third of the total distance between...

Let the total distance be $d$.

Time taken to cover the first one-third of the distance is $t_1 = \dfrac{d/3}{30} = \dfrac{d}{90}$.

Time taken to cover the remaining two-thirds of the distance is $t_2 = \dfrac{2d/3}{90} = \dfrac{2d}{270} = \dfrac{d}{135}$.

Total time taken for the entire journey is $t = t_1 + t_2 = \dfrac{d}{90} + \dfrac{d}{135} = \dfrac{3d + 2d}{270} = \dfrac{5d}{270} = \dfrac{d}{54}$.

Average speed is given by the total distance divided by the total time:

$v_{avg} = \dfrac{d}{t} = \dfrac{d}{d/54} = 54 \text{ km/h}$.

Answer: $54 \text{ km/h}$