An oil drop falling at constant velocity means the downward gravitational force equals the upward air resistance: $mg=F_r$. To make it rise at the same speed with an electric field, the net upward force must equal the same air resistance. Since the charge is negative, the electric force $F_e=qE$ acts opposite to the field direction. For upward motion, the electric force must be upward, so the field must be directed downward. The force balance is qE+Fr=mg. Since $F_r=mg$, we get $qE=2mg$. Substituting values: $E=\frac{2\times 1.6\times {10}^{-18}\times 10}{9.6\times {10}^{-19}}=\frac{1}{3}\times {10}^2\text{N/C}$ (mass converted to kg).

Final answer: the electric field is directed downward and the intensity of the electric field is $\frac{1}{3}$ × 10² N/C.