Two very large flat sheets face each other…

Step 1. One sheet. $E = \dfrac{\sigma}{2\varepsilon_0} = \dfrac{3.0 \times 10^{-6}\ \text{C/m}^2}{2 \times 8.85 \times 10^{-12}\ \text{C}^2/(\text{N m}^2)} = 1.69 \times 10^5\ \text{N/C}$, that is $169$ kN/C.

Step 2. Between the sheets. Both fields point from the positive sheet to the negative sheet. They add: $2 \times 169 = 339$ kN/C.

Step 3. Outside. The two fields point in opposite directions and have equal size. They cancel, and the field is $0$ kN/C.

Step 4. The electron. $F = eE = 1.6 \times 10^{-19}\ \text{C} \times 3.39 \times 10^5\ \text{N/C} = 5.42 \times 10^{-14}\ \text{N}$. The electron is pushed towards the positive sheet.