Notice that there's a degree of freedom here: the positioning of the paths. It doesn't matter how tightly packed together or where they are on the field, only that they are indeed vertical and horizontal respectively (in the aforementioned quantities, of n and n both in this case). So make your life easier and squish them all together at one corner, for instance the top left, for ease of visualization.

There is an square of path at the top left of side length n meters, then two rectangles adjoining it, one (100 - n) by n meters and the other (50 - n) by n meters. Adding the remaining area of 3096 m^(2) to the areas of these gives the total area of the field, 5000 m^(2). I.e. n^(2) + (100 - n)n + (50 - n)n + 3096 = 5000 or:

n^(2) - 150n + 1904 = 0

as an equivalent equation. Now solve for n however you like.