The image above is a rough representation of the question's drawing, including just the pertinent information, with the $10.6$ length coming from $19 - 8.4$. Note that due to the common $\angle ACE$, plus since $AE \parallel BD$, we also have that $\angle CBD = \angle CAE$ and $\angle CDB = \angle CEA$, with this giving that $\triangle BCD \sim \triangle ACE$. Thus, using the ratios of corresponding sides of a similar triangle, we get that
$$\cfrac{\left\lvert \, CD \, \right\rvert}{\left\lvert \, CB \, \right\rvert} = \cfrac{\left\lvert \, CE \, \right\rvert}{\left\lvert \, CA \, \right\rvert} \tag{1}\label{eq1}$$
Plugging in appropriate values from the diagram, we then get that
$$\cfrac{5.4}{10.6} = \cfrac{5.4 + \text{C}2}{19} \tag{2}\label{eq2}$$
Cross-multiplying and simplifying gives that
$$102.6 = 57.24 + 10.6 \; \text{C}2 \Rightarrow \text{C}2 = \cfrac{\left(102.6 - 57.24\right)}{10.6} = \cfrac{45.36}{10.6} \tag{3}\label{eq3}$$
The division creates a decimal starting with 4.279..., so you can decide yourself whether to use the exact value or an approximate decimal.
One interesting thing to note is that $\text{C}2 + 5.4$ is less than $10.6$, so the drawing is misleading as it makes it seem that this length should be longer.
Also, as you can see, you don't need to use the other information provided, i.e., $\text{C}1$ and the additional parallel lines. They were likely put in due to the context of a train & parallel lines, but possibly instead were used to try to confuse things, or perhaps to allow a different way to solve the problem (but I don't think any other method will be much, if any, simpler than what I've shown here).
