Circular Tracks

We can draw out a rough diagram as follows:

Let's say A takes 1 minute to cover one side of the park. In that 1 minute, C will make an entire loop and be at its starting point, and B will cross two edges. 

A and B will meet first at the 2 minute mark (at C's starting point), where C will also be. 
This would be the first instance of them all meeting together. 

The second meeting of A and B will again be at the same point, after 3 minutes. 

Then, they keep meeting at C's starting point every three minutes. 

So, the time taken for their third meeting would be 2+3+3 = 8 minutes. 

We initially assumed that C travels all three sides of the triangle in a single minute, meaning that in 8 minutes, he travelled a total of 24a in distance. 

Therefore, Option B is the correct answer. 

The relative speeds between A and B = 9kmph=2.5m/s The relative speeds between B and C = 18kmph=5m/s The relative speeds between A and C = 27kmph=7.5m/s

A and B meet after every 2000/(2.5) = 800 seconds. B and C meet after every 400 seconds and A& C meet after every 266.66 seconds. So, all three of them meet after every LCM(800/3, 400, 800) = 800 seconds

Alternate Solution : 

Relative speed of A and B = (22-13) = 9 km/hr = 2.5 m/s. Relative speed of B and C = (40-22) = 18 km/hr = 5 m/s. 
From this, we can see that A and B will meet only when the relative distance between them is a multiple of 2000m. Thus, time taken to cover a distance of 2000m = $$\frac{2000}{2.5} = 800 $$seconds. This means that A and B will meet after every interval of 800 seconds. 
We can also see that B and C will meet only when the relative distance between them is a multiple of 2000m. Thus, the time taken to cover a distance of 2000m= $$\frac{2000}{5} = 400 $$seconds. This means that B and C will meet after every interval of 400 seconds. 
Now, A,B,C will all meet at a point at a time which is the LCM of 800 and 400 seconds ie 800 seconds.