How many matches do you need?

If you want to accept this solution, you have to test the formulas. Here is a picture, what can help:

Let's do some formulas:

If there is a cuboid with side lengths a , b and c , we need $D_{3[a,b,c]}$ matches, where $D_{3[a,b,c]}=a(b+1)(c+1)+b(a+1)(c+1)+c(a+1)(b+1)$ (1. formula) , because:

We can divide this cuboid to $a*b$ rectangles $c+1$ times. Between this rectangles there are $c(a+1)(b+1)$ matches. So $D_{3}=(c+1)D_{2[a,b]}+c(a+1)(b+1)$ , where $D_{2[a, b]}=a(b+1)+b(a+1)$ (2.formula) , because:

This rectangles we can divide to lines with $a$ lenght, $b+1$ times. Between this lines there are $a+1$ lines with $b$ length. So $D_{2}=a(b+1)+b(a+1)$ .

If we have a flat surface and we want to build $a*b*1$ cubes for this, we need $D_{3[a,b]}$ pieces of matches, where $D_{3[a,b]}=D_{3[a,b,c]}-D_{2[a,b]}$ (3.formula) .

If there is a protruding cuboid and we want to place $a$ cubes next to this, we need $D_{3[a]}=a*3+2$ (4.formula) cubes.

First time we should search the whole part of the cube root of 2020, this is 12. Now $2020-12^3=292$ cubes left. We can place 2 times $12*12*1$ cubes to the sides of the big cube. After that $292-2*1*12*12=4$ cubes left. But we can place this cubes next to the second $12*12*1$ cube, because after the placing the first, the sides of the big cube are: 12, 12 and 13. Let's use the formulas:

The big cuboid: $D_{3[12, 12, 13]}$

The $12*12$ square on this side: $D_{3[12, 12]}$

And the last 4 squares: $D_{3[4]}$

The number of matches: $D_{3[12, 12, 13]}$ + $D_{3[12, 12]}$ + $D_{3[4]}=12(12+1)(13+1)+12(12+1)(13+1)+13(12+1)(12+1)+D_{3[1,12,12]}-D_{2[12,12]}+4*3+2=6565+1(12+1)(12+1)+12(1+1)(12+1)+12(1+1)(12+1)-12(12+1)-12(12+1)+14=6579+481=\boxed{7060}$