Solution:

Let's solve this briefly.

1. Charges and Positions:
- Charge on \( A = 8 \times 10^{-6} \, \text{C} \).
- Charge on \( B = -2 \times 10^{-6} \, \text{C} \).
- Distance between \( A \) and \( B = 20 \, \text{cm} \).

2. Equilibrium Condition for Third Charge \( Q \):
- For the third charge \( Q \) to experience zero net force, it must be positioned where the attractive force from \( A \) and the repulsive force from \( B \) on \( Q \) are equal in magnitude.
- Since \( |A| > |B| \), \( Q \) must be placed on the right side of \( B \) (further from \( A \)).

3. Distance Calculation:
- Let the distance of \( Q \) from \( B \) be \( x \).
- The distance of \( Q \) from \( A \) is then \( 20 + x \).

Using Coulomb's law, we set the magnitudes of the forces equal:
\[
\frac{k \cdot (8 \times 10^{-6}) \cdot Q}{(20 + x)^2} = \frac{k \cdot (2 \times 10^{-6}) \cdot Q}{x^2}
\]
Simplify by cancelling \( k \) and \( Q \):
\[
\frac{8}{(20 + x)^2} = \frac{2}{x^2}
\]
Cross-multiplying gives:
\[
8x^2 = 2(20 + x)^2
\]
Simplifying, we find \( x = 20 \, \text{cm} \).

Answer:
The third charge should be placed 20 cm to the right of \( B \) for zero net force.