All those particles are moving due to their mutual gravitational attraction.

The force between each masses are repulsive force.

On mass M at C, due to mass at D the repulsive force is F in the vertical direction.

On mass M at C, due to mass at B the repulsive force is F in the horizontal direction.

On mass M at C, due to mass at A the repulsive force is F'

Net force acting on particle at C,

= $$2F\,\cos \,{45^ \circ } + F'$$

Where $$F = {{G{M^2}} \over {{{\left( {\sqrt 2 R} \right)}^2}}}$$ and $$F' = {{G{M^2}} \over {4{R^2}}}$$

$$ \Rightarrow {{2 \times G{M^2}} \over {\sqrt 2 {{\left( {R\sqrt 2 } \right)}^2}}} + {{G{M^2}} \over {4{R^2}}}$$

$$ \Rightarrow {{G{M^2}} \over R^2}\left[ {{1 \over 4} + {1 \over {\sqrt 2 }}} \right] $$

This net force will balance by the centripetal force F_cp = $${{M{v^2}} \over R}$$

$$\therefore$$ $${{M{v^2}} \over R} = $$ $${{G{M^2}} \over R^2}\left[ {{1 \over 4} + {1 \over {\sqrt 2 }}} \right] $$

$$ \Rightarrow $$ $$v = \sqrt {{{Gm} \over R}\left( {{{\sqrt 2 + 4} \over {4\sqrt 2 }}} \right)} $$

$$ = {1 \over 2}\sqrt {{{Gm} \over R}\left( {1 + 2\sqrt 2 } \right)} $$