The electric field produced by a single-point charge is given by

where
k is the Coulomb's constant
q is the charge
r is the distance from the charge
To find the electric field at x=0.200 m, we need to find the electric field produced by each charge at that point, and then find their resultant.
1) The first charge is
, and it is located at x=0, so its distance from the point x=0.200 m is

Therefore, the electric field is

And since the charge is negative, the direction of the field is toward the charge, so toward negative x direction.
2) The second charge is
and it is located at x=0.800 m, so its distance from the point is

Therefore, the electric field is

And since the charge is negative, the direction of the field is toward the charge, so toward positive x-direction.
3) The total electric field at x=0.200 m will be given by the difference between the two fields (because they are in opposite directions). Taking the x-positive direction as positive direction, we have

and the sign tells us that the field is directed toward negative x-direction.