Firstly, from the geometry you can see that (putting A and B at the top of the square and C and D at the bottom of the square) all of the points will have the same magnitude effect of keQ/(s/sqrt(2))2 where s is the side of the square (i.e. 2keQ/s2). Secondly, the x components of E fields from A and B will balance at the center, as will the x components of the E fields from C and D. Thirdly, the y components will all result in a pull in the - y direction (C and D pulling down and A and B pushing down). The y component magnitude from any of the points is (sqrt(2)/2) (2keQ/s2) = sqrt(2)keQ/s2 and because each point contributes: |Ey | = 4sqrt(2) keQ/s2
When you plug in you must use: Q = 5 x 10-6C and s = .01m in order to obtain N/C. Please check my math as there are lots of sqrt(2) twists.