Hmm,
The initial speed is not 0, as the cricket wouldn't go up for starters. In order to solve this question, you should make abstraction (ie forget about) the acceleration of the cricket, that is the time it takes for him to go from rest to full take-off speed.
It does not take 1/3rd of the time to travel to 1/3rd of the height, as in the beginning the cricket is going very fast, and is gradually loosing speed, therefore requiring less time to do the first part of the jump than for the last part of it.
So, how to tackle the question? Assume position axis is pointing up and the origin is on ground level.
Assume the cricket has a start speed of v0. It will go up, and the speed will diminish until it's zero, at which moment it starts falling down again.
The speed will diminish with g m/s per second. That is, v = vo - g.t
During a given time t, the distance traveled is x=x0+v0t-at²/2
These formula's should look quite familiar to you if you have to solve the question. There might be a difference in signs however, depending how you define the axis.
The way to go is to find out first how much time it takes to get to the top of the jump. At that moment, you know a lot of parameters for both equations, enough to conveniently plug one into the other.
From there, the rest should be simple.
See if you get further this way. If not, show us your way of reasoning, and we'll hint you further. Sorry for not just typing out the solution, but keep in mind we won't be there during tests either...
Good luck !
Leo