1. For an uncharged oil drop in Millikan’s experiment, when it falls steadily, the viscous force balances its weight (since buoyancy is neglected).

   So, $F_v = mg$

2. Mass of the drop:

   Radius, $r = 2.0 \times 10^{-5}\ \text{m}$

   Density, $\rho = 1.2 \times 10^3\ \text{kg m}^{-3}$

   Volume of the spherical drop, $V = \frac{4}{3}\pi r^3$

   Hence mass, $m = \rho V = \rho \cdot \frac{4}{3}\pi r^3$

3. Compute $r^3$:

   $r^3 = (2.0 \times 10^{-5})^3 = 8.0 \times 10^{-15}\ \text{m}^3$

4. Compute mass:

   $m = 1.2 \times 10^3 \times \frac{4}{3}\pi \times 8.0 \times 10^{-15}$

   $m = 1.2 \times 10^3 \times \frac{32}{3}\pi \times 10^{-15}$

   Using $\frac{32}{3}\pi \approx 33.51$,

   $m \approx 1.2 \times 10^3 \times 33.51 \times 10^{-15}$

   $m \approx 4.02 \times 10^{-11}\ \text{kg}$

5. Compute viscous force:

   $F_v = mg = 4.02 \times 10^{-11} \times 9.8$

   $F_v \approx 3.94 \times 10^{-10}\ \text{N}$

   So, $F_v \approx 3.9 \times 10^{-10}\ \text{N}$

6. Match with options:

   The correct option is: $\boxed{\text{B: } 3.9 \times 10^{-10}\ \text{N}}$

Note: The given viscosity is not needed here because the question asks the viscous force on the drop, and at terminal speed for an uncharged drop, viscous force equals weight.