Calculate the magnitude and direction

Of an electric field using a small test charge located .30m to the right of a charge of -3.0x10^-6

just use formula

E = k*q/r^2

now k = 9*10^9 which is 1/(4*pi*epsilon)

E = 9*10^9*(-3.0*10^6)/(3*10^-1)^2

E = 9*10^9*(-3.0*10^-6)/(9*10^-2)

alot of the numbers cancel

E = 1*10^9*(-3.0*10^-6)*10^2

the last exponent I inverted the 1/10^-2 to make it 10^2

E = -3.0*10^3*10^2
E = -3.0*10^5

this means it is a magnitude of 3.0*10^5

and the direction is toward the original (not test) charge because its negative..
this means it points left
hope that helps