Answer
$E = (-6.39\times 10^5~N/C)~\hat{i}$
Work Step by Step
Note that the midway point between the particles is at the point $x = 13.5~cm$
We can find the magnitude of the electric field at the midpoint due to particle 1:
$E = \frac{\vert q \vert}{4\pi~\epsilon_0~r^2}$
$E = \frac{(2.00\times 10^{-7}~C)}{(4\pi)(8.854\times 10^{-12}~C^2/N~m^2)~(0.075~m)^2}$
$E = 3.1956\times 10^5~N/C$
Since particle 1 has a negative charge, the electric field at the midpoint due to particle 1 points toward particle 1 in the negative x direction.
We can express this electric field in unit-vector notation:
$E_1 = (-3.1956\times 10^5~N/C)~\hat{i}$
We can find the magnitude of the electric field at the midpoint due to particle 2:
$E = \frac{\vert q \vert}{4\pi~\epsilon_0~r^2}$
$E = \frac{(2.00\times 10^{-7}~C)}{(4\pi)(8.854\times 10^{-12}~C^2/N~m^2)~(0.075~m)^2}$
$E = 3.1956\times 10^5~N/C$
Since particle 2 has a positive charge, the electric field at the midpoint due to particle 2 points away from particle 2 in the negative x direction.
We can express this electric field in unit-vector notation:
$E_2 = (-3.1956\times 10^5~N/C)~\hat{i}$
We can find the net electric field:
$E = E_1+E_2$
$E = (-3.1956\times 10^5~N/C)~\hat{i}+(-3.1956\times 10^5~N/C)~\hat{i}$
$E = (-6.39\times 10^5~N/C)~\hat{i}$
