Answer
$6.30\times10^6\;N/C\;upward$.
Work Step by Step
The electric field created by a positive charge points directly away from it.
Use equation 16–4a to calculate the electric field.
$$E=k\frac{Q}{r^2}$$
$$=(8.99\times10^9\frac{N\cdot m^2}{C^2})\frac{33\times10^{-6}C}{(21.7\times10^{-2}m)^2}$$
$$= 6.30\times10^6\;N/C\;upward$$
