Two point charges A and B of magnitude + 8 × 10 - 6 C and - 8 × 10 - 6 C respectively are placed…

Direction of electric field at mid-point due to both charge will be in same direction, which is towards right.

Therefore, $E$ at mid-point $=\frac{kq}{{\left(\frac{d}{2}\right)}^2}+\frac{kq}{{\left(\frac{d}{2}\right)}^2}$ 

$\Rightarrow 6.4\times {10}^4=2\times \left[9\times {10}^9\times \frac{\left(8\times {10}^{-6}\right)}{d^2}\times 4\right]$

$\Rightarrow d=3 \mathrm{m}$