Four particles of masses $m, 2m, 3m$ and $4m$ are kept in sequence at the corners of a square of side $a$. The magnitude of gravitational force acting on a particle of mass $m$ placed at the centre of the square will be:

$$\begin{gathered} F_1=\frac{Gmm}{{\left({\displaystyle \raisebox{1ex}{$a$}\!\left/ \!\raisebox{-1ex}{$\sqrt{2}$}\right.}\right)}^2}, F_2=\frac{Gm2m}{{\left({\displaystyle \raisebox{1ex}{$a$}\!\left/ \!\raisebox{-1ex}{$\sqrt{2}$}\right.}\right)}^2} \\ {F}_3=\frac{Gm3m}{{\left({\displaystyle \raisebox{1ex}{$a$}\!\left/ \!\raisebox{-1ex}{$\sqrt{2}$}\right.}\right)}^2}, F_4=\frac{Gm4m}{{\left({\displaystyle \raisebox{1ex}{$a$}\!\left/ \!\raisebox{-1ex}{$\sqrt{2}$}\right.}\right)}^2} \end{gathered}$$
So resultant of forces will be:
 

$$\begin{gathered} F\text{'}=\frac{Gm4m}{{\left({\displaystyle \raisebox{1ex}{$a$}\!\left/ \!\raisebox{-1ex}{$\sqrt{2}$}\right.}\right)}^2}-\frac{Gm2m}{{\left({\displaystyle \raisebox{1ex}{$a$}\!\left/ \!\raisebox{-1ex}{$\sqrt{2}$}\right.}\right)}^2}, F\text{'}\text{'}=\frac{Gm3m}{{\left({\displaystyle \raisebox{1ex}{$a$}\!\left/ \!\raisebox{-1ex}{$\sqrt{2}$}\right.}\right)}^2}-\frac{Gmm}{{\left({\displaystyle \raisebox{1ex}{$a$}\!\left/ \!\raisebox{-1ex}{$\sqrt{2}$}\right.}\right)}^2} \\ F\text{'}=\frac{2G{m}^2}{{\displaystyle \raisebox{1ex}{$a^2$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}, F\text{'}\text{'}=\frac{2G{m}^2}{{\displaystyle \raisebox{1ex}{$a^2$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}} \\ F\text{'}=\frac{4G{m}^2}{a^2}. F\text{'}\text{'}=\frac{4G{m}^2}{a^2} \end{gathered}$$
Now they are at $90°$
So resultant force$=4\sqrt{2}\frac{G{m}^2}{a^2}$