Answer:
  E = k q / a²   (1.3535) (- i ^ + j ^)
   E = k q / a²  1.914  ,      θ’= 135
Explanation:
For this exercise we will use Newton's second law where we must add as vectors
         E_total = E₁₂ i ^ + E₁₄ j ^ + E₁₃
Let's look for the value of each term
On the x axis
        E₁₂ = k q / a²
On the y axis
        E₁₄ = k q / a²
For the charge in the opposite corner we look for the distance
         d = √ (a² + a²) = a √2
let's look for the field
       E₁₃ = k q / d²
       E₁₃ = k q / 2a²
let's use trigonometry to find the two components of this field
        cos 45 = E₁₃ₓ / E₁₃
        E₁₃ₓ = E₁₃ cos 45
        
        sin 45 = E_{13y} / E₁₃
        E_{13y} = E₁₃ sin 45
        E₁₃ₓ = k q / 2a²  cos 45
        E_{13y} = k q / 2a²  sin 45
let's find each component of the electric field
X axis
       Eₓ = -E₁₂ - E₁₃ₓ
       Eₓ = - k q / a² - k q / 2a² cos 45
       Eₓ = - k q / a² (1 + cos 45/2)
       cos 45 = sin 45 = 0.707
       Eₓ = - k q / a²   (1 + 0.707 / 2)
       Eₓ = - k q / a²    (1.3535)
Y axis  
       E_y = E₁₄ + E_{13y}
        E_y = k q / a² + k q / 2a²     sin 45
        E_y = k q / a² (1 + sin 45/2)
        E_y = k q / a²       (1.3535)
we can give the results in two ways
        E = k q / a²   (1.3535) (- i ^ + j ^)
In modulus and angle form, let's use Pythagoras' theorem for the angle
        E = √ (Eₓ² + E_y²)
         E = k q / a²    1.3535 √2
         E = k q / a²     1.914
we use trigonometry for the angle
         tan θ = E_y / Eₓ
          θ = tan⁻¹  (E_y / Eₓ)
          θ = tan⁻¹ (1 / -1)
          θ = 45
in the third quadrant, if we measure the angle of the positive side of the x-axis
            θ‘= 90 + 45
            θ’= 135