Answer
The magnitude of electric field will be $E=3.2\times10^{2}N/C$
and direction will be away from the point charge.
Work Step by Step
magnitude of the electric field due to a point charge is
$E= \frac{kq}{r^2}$
given that $q=+2pC=+2\times10^{-12}C$
$r=0.75cm=0.75\times10^{-2}m=7.5\times10^{-3}m$
$k=9.0\times10^{9}N.m^2/C^2$
so magnitude of electric field will be
$E= \frac{9.0\times10^{9}N.m^2/C^2\times 2\times10^{-12}C}
{(7.5\times10^{-3}m)^2} =0.32\times10^{3}N/C=3.2\times10^{2}N/C$
since it is a positive charge its direction will be away from the charge at any point of space.
