A

Step 1: Understanding the electric field due to infinite sheets
For an infinite charged sheet with surface charge density σ, the electric field produced is given by:
$$E = \frac{\sigma}{2\epsilon_0}$$
where ${\epsilon_0}$ is the permittivity of free space.

Step 2: Electric field directions
1. For the sheet with charge density σ (at point A), the electric field points away from the sheet. Hence, it contributes an electric field going upwards (positive y-direction).
2. For the sheet with charge density 2σ (at point B), it also points away from the sheet, adding another electric field upwards above point B.
3. For the sheet with charge density -4σ (at point C), the electric field points towards the sheet (downwards), contributing to the electric field at point D.

Step 3: Calculating electric fields at points A, B, C, and D
1. **At point A**:
Only the sheet with charge density σ affects point A, giving:
$$E_A = \frac{\sigma}{2\epsilon_0} (upward)$$

2. **At point B**:
Both σ and 2σ contribute upwards and have magnitudes:
$$E_B = \frac{\sigma}{2\epsilon_0} + \frac{2\sigma}{2\epsilon_0} = \frac{3\sigma}{2\epsilon_0} (upward)$$

3. **At point C**:
Here, conduct for sheet σ (upward) and 2σ (upward) as well as -4σ (downward):
$$E_C = \frac{\sigma}{2\epsilon_0} + \frac{2\sigma}{2\epsilon_0} - \frac{4\sigma}{2\epsilon_0} = -\frac{\sigma}{2\epsilon_0} (downward)$$

4. **At point D**:
The fields from σ, 2σ and -4σ sheets affect point D as follows:
$$E_D = \frac{\sigma}{2\epsilon_0} + \frac{2\sigma}{2\epsilon_0} - \frac{4\sigma}{2\epsilon_0} = -\frac{\sigma}{2\epsilon_0} (downward)$$

Summary of Electric Fields:
- E_A: upward, $$\frac{\sigma}{2\epsilon_0}$$
- E_B: upward, $$\frac{3\sigma}{2\epsilon_0}$$
- E_C: downward, $$-\frac{\sigma}{2\epsilon_0}$$
- E_D: downward, $$-\frac{\sigma}{2\epsilon_0}$$

Thus, based on point A, B being the maximum (upward), and point C, D being the negative field (downward). Therefore, the correct answer is A.