Four balls of 8.0 kg each sit at the corners of a square of side 0.40 m…

Step 1. Neighbours. Each pulls $\dfrac{Gm^2}{a^2} = 6.67 \times 10^{-11}\ \text{N m}^2/\text{kg}^2 \times \dfrac{64\ \text{kg}^2}{0.16\ \text{m}^2} = 2.668 \times 10^{-8}\ \text{N}$.

Step 2. Diagonal. $\dfrac{Gm^2}{2a^2} = 1.334 \times 10^{-8}\ \text{N}$.

Step 3. Add. The two side pulls give $2 \times 2.668 \times 10^{-8}\ \text{N} \times \cos 45^\circ = 3.773 \times 10^{-8}\ \text{N}$ along the diagonal; with the third, $5.11 \times 10^{-8}\ \text{N}$.