Answer
$1.16\times10^5\;N/C\;south$.
Work Step by Step
Use equation 16–3 to calculate the direction and magnitude of the electric field.
$$\vec{E}=\frac{\vec{F}}{q}$$
$$\vec{E}=\frac{1.86\times10^{-14}\;N \;south}{1.602\times10^{-19}C }$$
$$= 1.16\times10^5\;N/C\;south$$