Calculate the angular velocity of the milk carton

No. You need to use the moment of inertia about the corner of the carton and you need to find the correct height of the center of mass when tipping occurs.

 

That is true, but since no mention is made in the problem about liquid in the carton, we may assume that it is empty. It could an empty carton on a cafeteria conveyor belt on its way to the trash disposal area.

On edit: Of course, the moment of inertia of an empty carton is more complicated to calculate.

 

It's also an issue for cartons that are full. The milk would tend to rotate internally in the opposite direction.
But an earlier part of the question says to ignore any sloshing around https://www.physicsforums.com/threads/calculate-the-maximum-acceleration-of-a-milk-carton.911818/.

There is another complication. Is it easily shown that the carton does not lose contact with the ground?

 

Why are you dividing the 0.195 by 2 but not the 0.07?

 

And why do you think the CM is at height ##\frac{0.195}{2}~m## just before the carton tips over? That's its height when the carton sits flat on the conveyor belt.

 

Yes, but it is important to understand why. Start with ##I_{CM}=\frac{1}{12}M(l^2+a^2)## and apply the parallel axes theorem. Try using symbols, not numbers. What do you get?

 

I assume you mean Md², not M+d².
What is d² in terms of l and a?

 

So what do you get now for the vertical displacement of the mass centre?