I would like to measure the volumes of these three bags because the given values don't make sense (Feel free to guess the volume/rank them by size in the comments. I will update the question with the true values in a couple of days). One way, I imagine, to do that would be to get a lot of tennis balls or similar and see how many that fit into each bag. That way you can get a pretty exact relative size (if one bag can hold 10 balls and another 15 balls, the latter is 1,5 times bigger) but that doesn't answer the question "what is the actual volume in litres?" because there will be a lot of empty space between the balls.
Is it possible to give a good estimate of how much space there are between the balls if they are packed in a perfect way? The exact size of the balls is known.
(Are there other ways than balls to estimate the size in this situation? The bags are not watertight, and besides, if you load them with 50 to 100 kg of water I think they will break.)
Let's say all your tennis balls have the exact same volume $V$, and they don't get squished in the bag.
If you pour the tennis balls in the bag randomly roughly $40\%$ of the bag will be empty. If you shake the bag and add more balls it will be approximately $36\%$ empty. Source:[https://en.wikipedia.org/wiki/Random_close_pack]
So, lets suppose that the bag is full of tennis balls, and $36\%$ empty. In order words, the tennis balls only account for $64\%$ of the total volume.
Then, if you add $n$ balls the total volume of the bag is $nV*\frac{100}{64}$.
This is a bit awkward because you have to try add the balls to the bag as randomly as possible. If you think you are able to pack them efficiently, the volume will be $nV*\frac{100}{76}$. This is because the densest packing of spheres will leave $24\%$ empty space. So the real answer will be somewhere in the range $\frac{100nV}{76}$ to $\frac{100nV}{64}$.