B

Step 1: Identify the weights of the blocks:
- Block 1 (top): 2 kg
- Block 2 (middle): 2 kg
- Block 3 (bottom): 4 kg.

Step 2: Calculate the normal forces acting on each block.
The weight of Block 1 is: \( W_1 = m_1 \cdot g = 2 \times 10 = 20 \, \text{N} \)
The weight of Block 2 is: \( W_2 = m_2 \cdot g = 2 \times 10 = 20 \, \text{N} \)
The weight of Block 3 is: \( W_3 = m_3 \cdot g = 4 \times 10 = 40 \, \text{N} \)

Step 3: Determine the friction forces:
- Top block (Block 1) on Block 2: \( F_{1} = \mu_1 \cdot W_1 = 0.1 \cdot 20 = 2 \, \text{N} \)
- Middle block (Block 2) on Block 3: \( F_{2} = \mu_2 \cdot W_2 = 0.6 \cdot 20 = 12 \, \text{N} \)

Step 4: Calculate the total friction force that must be overcome to start the sliding of Block 1 over Block 2.
When Block 1 starts sliding over Block 2, the applied force \( F \) must overcome the static friction force, which is limited to the friction force between Block 2 and Block 3:
Therefore, for Block 1 to start sliding, it must overcome the friction force \( F_{2} \): \( F = F_2 = 12 \, \text{N} \)
However, we must also overcome the friction between Block 1 and Block 2, which adds up:

Total force required to start sliding = \( F_{1} + F_{2} = 2 + 12 = 14 \, \text{N} \)

Therefore, the minimum horizontal force \( F \) must be: \( F = 14 \, \text{N} \). Thus, option B is correct.