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