Solution:

To make the oil drop move upward with the same terminal velocity it had while moving downward, the electric force upward must balance the gravitational force downward.

Solution Steps:

1. Given Data:
- Mass of the drop, \( m = 1.6 \times 10^{-12} \, \text{g} = 1.6 \times 10^{-15} \, \text{kg} \)
- Charge on the drop, \( q = 6 \times e = 6 \times 1.6 \times 10^{-19} \, \text{C} = 9.6 \times 10^{-19} \, \text{C} \)
- Gravitational force, \( F_{\text{gravity}} = mg \), where \( g = 9.8 \, \text{m/s}^2 \)

2. Gravitational Force:
\[
F_{\text{gravity}} = (1.6 \times 10^{-15}) \times 9.8 = 1.568 \times 10^{-14} \, \text{N}
\]

3. Electric Field Required:
- To counteract this gravitational force, the electric force \( F_{\text{electric}} = qE \) must equal \( F_{\text{gravity}} \).
\[
qE = F_{\text{gravity}}
\]
\[
E = \frac{F_{\text{gravity}}}{q} = \frac{1.568 \times 10^{-14}}{9.6 \times 10^{-19}} = 3.3 \times 10^4 \, \text{N/C}
\]

Conclusion:
The required electric field magnitude is \( 3.3 \times 10^4 \, \text{N/C} \).