Let one edge of the small corner squares be x. The volume of the box is (20-2x)(20-2x)x which is 4x^3-80x^2+400x. As you increase x by integers the volume of the box increase until you get to 4 when it starts decreasing.
As it’s a cubic there will be a maximum and a minimum for x which will give the largest volume and smallest volume. To work these out I differentiated 4x^3-80x^2+400x to get 12x^2-160x+400. This equation gives the gradient of the cubic at points of x and the minimum and maximum have gradients of 0. Therefore 12x^2-160x+400=0. I used the quadratic formula to get x=10 and 10/3. I put 10 back into the original cubic and it gave me the answer 0. This is because when x=10 there is no paper as 10x2=20. Therefore 10/3 is the maximum value of x and when this is put in the cubic the volume is 592.592 recurring so 593 to 3 significant figures.
I did this method again with 40cm of paper which gave me an x of 20/3 and volume of 4740.740 recurring. I also did it with 65cm and got an x of 65/6 and volume of 20342.592 recurring. From this the pattern for the maximum x is the full edge of the paper divided by 6. I also found that the volume pattern is the previous volume times the multiplier of the edge cubed. For example, the volume of the 40cm piece of paper is 592.592 (volume of 20cm piece) times 2 cubed (8) which is 4740.740. The volume of the 65cm piece is 4740.740 times 1.625 (65/40) cubed (4.29 (3sf)) which is 20342.592.