An electron is released from rest in a uniform electric field accelerates vertically upward. Acceleration of e is notice to have a magnitude of $ 1.6 \times 10^{12} m/s^2$.The direction and magnitude of electric field is :

$\text{Given} e = - 1.6 \times 10^{-19} C$

$\text{mass of e} = 9.10 \times 10^{-31}kg$ 

$\text{Magnitude of gravitational are g} = 10 m/s^2$

9.10 N/C in downward direction

Since charge on electron is negative so its acceleration is opposite to electric field.

$ \Rightarrow \text{ Direction of Electric Field is in Downward direction}$

$ \text{Magnitude of Electric Field is } E = \frac{ma}{q} = \frac{9.10\times 10^{-31} \times 1.6\times 10^{12}}{1.6\times 10^{-19}} = 9.1N/C$