According to the station clocks, two babies are born at the exact...

Let the ground frame be $S$ and the train frame be $S'$.

Let Agra be at the origin $x = 0$ and Kanpur be at $x = L$, where $L > 0$.

The train is travelling from Agra to Kanpur, so its velocity $v$ is in the positive $x$-direction.

The two births are simultaneous in the ground frame, so $t_A = t_K = 0$.

Using the Lorentz transformation for time, t' = \gamma \left( t -\dfrac{vx}{c^2} \right).

For the baby born in Agra: t'_A = \gamma \left( 0 -\dfrac{v(0)}{c^2} \right) =$0$.

For the baby born in Kanpur: t'_K = \gamma \left( 0 -\dfrac{vL}{c^2} \right) = -\dfrac{\gamma vL}{c^2}.

Since $v > 0$ and $L > 0$, we have $t'_K < t'_A$.

This means that in the train's reference frame, the baby in Kanpur was born earlier than the baby in Agra. Therefore, the Kanpur baby is elder.