 Solution

Choose $z=0$ at the sheet, positive $z$ upwards, so the uniformly charged slab occupies $-d<z<0$. Use the usual isolated-plane superposition convention, with no imposed uniform background field. The slab alone produces $+\rho d/(2{ϵ}_0)\hat{\mathbf{z}}$ above it, while the negatively charged sheet produces $-\sigma /(2{ϵ}_0)\hat{\mathbf{z}}$ there. ConsequentlySince $\rho d>\sigma$, the field is upward, away from the net positive charge distribution, with magnitude $(\rho d-\sigma )/(2{ϵ}_0)$. The equal halves in the separate infinite-sheet fields follow from their reflection symmetry before superposition.