Answer
$5.97\times10^{-10}N/C \;south$.
Work Step by Step
Assume that this is the only force acting on the electron. Apply Newton’s second law, and use equation 16–3 to calculate the magnitude of the field.
$$\vec{E}=\frac{\vec{F}}{q}$$
$$\vec{F}=q\vec{E}=m\vec{a}$$
$$\vec{E}=\frac{m\vec{a}}{q}=\frac{(9.11\times10^{-31}kg)(105\;m/s^2\;north)}{-1.602\times10^{-19}C}$$
$$= 5.97\times10^{-10}N/C \;south$$
