Charge −Q is located at (d, 0). The distance from the origin is d. Since the charge is negative, the electric field $\mathrm{vec}{\mathit{E}}_1$ at the origin points toward the charge, i.e., along $+\stackrel{\wedge }{\mathit{i}}:$
$\mathrm{vec}{\mathit{E}}_1=\frac{1}{4\mathit{\pi }{\mathit{E}}_{\mathit{o}}}\mathit{Q}\frac{{\mathit{d}}^2}{\mathit{i}}$
Charge +Q is located at (0, d). The distance is d. Since it is positive, the electric field $\mathrm{vec}{\mathit{E}}_2$ at the origin points away from the charge, i.e., along $-\stackrel{\wedge }{\mathit{j}}:$
$\mathrm{vec}{\mathit{E}}_2=\frac{1}{4\mathit{\pi }{\mathit{E}}_{\mathit{o}}}\frac{\mathit{Q}}{{\mathit{d}}^2}\left(-\stackrel{\wedge }{\mathit{j}}\right)$
The resultant electric field is the vector sum:
$\overrightarrow{\mathit{E}}=\mathrm{vec}{\mathit{E}}_1+\mathrm{vec}{\mathit{E}}_2$
$\overrightarrow{\mathit{E}}=\frac{1}{4\mathit{\pi }{\mathit{E}}_{\mathit{o}}}\mathit{Q}\frac{{\mathit{d}}^2}{\mathit{i}}$ - $\frac{1}{4\mathit{\pi }{\mathit{E}}_{\mathit{o}}}\frac{\mathit{Q}}{{\mathit{d}}^2}\stackrel{\wedge }{\mathit{j}}$
$\overrightarrow{\mathit{E}}=$$\frac{1}{4\mathit{\pi }{\mathit{E}}_{\mathit{o}}}\frac{\sqrt{2}\mathit{Q}}{{\mathit{d}}^2}\left(\stackrel{\wedge }{\mathit{i}}-\stackrel{\wedge }{\mathit{j}}\right)$
The magnitude would be $\frac{\mathit{Q}\sqrt{2}}{4\mathit{\pi }{\mathit{E}}_{\mathit{o}}{\mathit{d}}^2}$, but the vector expression matches the required form.
The electric field is $\frac{1}{4\mathit{\pi }{\mathit{E}}_{\mathit{o}}}\frac{\sqrt{2}\mathit{Q}}{{\mathit{d}}^2}\left(\stackrel{\wedge }{\mathit{i}}-\stackrel{\wedge }{\mathit{j}}\right)$.