#### Answer

The last car to the west pulls on the rest of the train with a force of $2.0\times 10^5~N$ (directed to the west).

#### Work Step by Step

Let $M$ be the mass of each train car. We can find the acceleration of the train:
$a = \frac{F}{10~M} = \frac{2.0\times 10^6~N}{10~M} = \frac{2.0\times 10^5~N}{M}$
We can find the force $F_p$ with which the ninth car pulls on the last car:
$F_p = Ma = M~(\frac{2.0\times 10^5~N}{M}) = 2.0\times 10^5~N$
By Newton's third law, the last car pulls on the ninth car with a force of the same magnitude.
The last car to the west pulls on the rest of the train with a force of $2.0\times 10^5~N$ (directed to the west).