Trains

Speed of the faster train = $$\frac{160}{12}=\frac{40}{3}\ $$ m/s

Speed of the slower train = $$\frac{40}{3}-\left(6\times\ \frac{5}{18}\right)=\frac{35}{3}$$ m/s

Sum of speeds (when the trains travel towards each other) = $$\frac{40}{3}+\frac{35}{3}=25$$ m/s

Let the slower train be $$x$$ metres long; then: $$\frac{160+x}{25}=14$$

On solving, $$x=190\ m$$

Considering the length of train A = La, length of train B = Lb.

The speed of train A be 5*x, speed of train B be 3*x.

From the information provided :

The front end of A crossed the rear end of B 46 seconds after the front ends of the trains had crossed each other.

In this case, train A traveled a distance equivalent to the length of train B which is Lb at a speed of 5*x+3*x = 8*x because both the trains are traveling in the opposite direction.

Hence (8*x)*(46) = Lb.

In the information provided :

It took another 69 seconds for the rear ends of the trains to cross each other.

In the next 69 seconds 

The train B traveled a distance equivalent to the length of train A in this 69 seconds.

Hence (8*x)*(69) = La.

La/Lb = 69/46 = 3/2 = 3 : 2

Speed in m/s = 72/3.6 m/s = 20m/s Total distance = 1440 metres Total time taken = 1440/20 = 72 seconds.