Four identical particles of mass m are kept at the four corners of...

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

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

Therefore, ${F}_{net} = \sqrt{2}\dfrac{G{m}^{2}}{{a}^{2}} + \dfrac{G{m}^{2}}{2{a}^{2}}$

Comparing with the expression given in the question, we get

\left(\dfrac{2\sqrt{2} + 1}{32}\right)\dfrac{G{m}^{2}}{{L}^{2}} = \dfrac{G{m}^{2}}{{a}^{2}}\left(\dfrac{2\sqrt{2} + 1}{2}\right)$</p><p>$⇒ a = 4L