A

Step 1: Calculate the electric potential (V) at points A, B, and C due to the charge at the origin.
The electric potential due to a point charge is given by the formula:
$$ V = \frac{kQ}{r} $$
where $k \approx 8.99 \times 10^9 \ \text{N m}^2/\text{C}^2$ is Coulomb's constant, $Q$ is the charge at the origin (8 mC = 8 \times 10^{-3} C), and $r$ is the distance from the charge to the point where the potential is being calculated.

Step 2: Calculate the distance r for points A, B, and C.
- For point A (0, 0, 0.03 m):
$$ r_A = \sqrt{0^2 + 0^2 + (0.03)^2} = 0.03 \ m $$
- For point B (0, 0.04 m, 0):
$$ r_B = \sqrt{0^2 + (0.04)^2 + 0^2} = 0.04 \ m $$
- For point C (0, 0.06 m, 0.09 m):
$$ r_C = \sqrt{0^2 + (0.06)^2 + (0.09)^2} = \sqrt{0.0036 + 0.0081} = \sqrt{0.0117} \approx 0.1081 \ m $$

Step 3: Calculate the potentials.
- At point A:
$$ V_A = \frac{(8.99 \times 10^9)(8 \times 10^{-3})}{0.03} \approx \frac{71.92 \times 10^6}{0.03} = 2.3973 \times 10^9 \ V $$
- At point B:
$$ V_B = \frac{(8.99 \times 10^9)(8 \times 10^{-3})}{0.04} \approx \frac{71.92 \times 10^6}{0.04} = 1.798 \times 10^9 \ V $$
- At point C:
$$ V_C = \frac{(8.99 \times 10^9)(8 \times 10^{-3})}{0.1081} \approx \frac{71.92 \times 10^6}{0.1081} \approx 665.5 \times 10^6 \ V $$

Step 4: Calculate the work done (W) in moving the charge from A to B via C.
Work done in moving a charge is given by:
$$ W = q(V_f - V_i) $$
where $q = -2 \times 10^{-9} C$, $V_i$ is the initial potential, and $V_f$ is the final potential.
- Moving from A to C:
$$ W_{AC} = -2 \times 10^{-9}(V_C - V_A) = -2 \times 10^{-9}(665.5 \times 10^6 - 2.3973 \times 10^9) \approx -2 \times 10^{-9}(-1.7318 \times 10^9) \approx 3.4636 \times 10^{-9} J $$
- Moving from C to B:
$$ W_{CB} = -2 \times 10^{-9}(V_B - V_C) = -2 \times 10^{-9}(1.798 \times 10^9 - 665.5 \times 10^6) \approx -2 \times 10^{-9}(1.1325 \times 10^9) \approx -2.265 \times 10^{-9} J $$
Step 5: Total work done (W_total) from A to B:
$$ W_{total} = W_{AC} + W_{CB} \approx 3.4636 \times 10^{-9} - 2.265 \times 10^{-9} \approx 1.1986 \times 10^{-9} J $$
Therefore, the work done by the external agent is approximately 1.1986 nJ.