The figure above shows the railway tracks between three

$$Length\ of\ BC\ =\ h^2=O^2+a^2\ <pythagoras>$$
$$\ h^2=\left(100^2\right)km+\left(240^2\right)km$$
$$\ h^2=10000+57600=67600$$
$$\ h=\sqrt{67600}$$
$$\ h=260km$$

Length of BC = 260km
Distance between A to B via C= 240+260 = 500km
First train leave 1/2 hr (30 minutes) early
Second train leaves 1/2 hr (30 minutes) late

First late leaves at 08:30 am
Second train leaves at 09:30 am
When second train leaves the first train, the first train has already traveled for 1 hour and has covered a distance of 20km with speed of 20 km per hour.
Distance between first train from the second train is 500km - 20km = 480km

Time required for the 2 trains to meet = (distance between them) / speed
$$\frac{480}{20+25}=\frac{480}{45}=\frac{32}{3}$$
$$=10\ \frac{2}{3}=10hours\ and\ 40\ \min utes$$
9:30am + 10 hours 40 mins = 8:10 pm

Answer is B