Four identical particles of mass m are kept at the four corners of a square. If the gravitational force…

Resultant force, $F_{net}=\sqrt{2}F+F\text{'}$, where

$F=\frac{G{m}^2}{a^2}$ and $F\text{'}=\frac{G{m}^2}{{\left(\sqrt{2}a\right)}^2}$

Therefore, $F_{net}=\sqrt{2}\frac{G{m}^2}{a^2}+\frac{G{m}^2}{2a^2}$

Comparing with the expression given in the question, we get

$\left(\frac{2\sqrt{2}+1}{32}\right)\frac{G{m}^2}{L^2}=\frac{{\displaystyle G{m}^2}}{{\displaystyle a^2}}\left(\frac{{\displaystyle 2\sqrt{2}+1}}{{\displaystyle 2}}\right)$

$\Rightarrow a=4L$