Net force on mass M at position B towards centre of circle is

$${F_{BO\,net}} = {F_{BD}} + {F_{BA}}\sin 45^\circ + {F_{BC}}\sin 45^\circ $$

$$ = {{G{M^2}} \over {{{(\sqrt 2 a)}^2}}} + {{G{M^2}} \over {{a^2}}}\left( {{1 \over {\sqrt 2 }}} \right) + {{G{M^2}} \over {{a^2}}}\left( {{1 \over {\sqrt 2 }}} \right)$$

[where, diagonal length BD is $$\sqrt2$$a]

$$ = {{G{M^2}} \over {2{a^2}}} + {{G{M^2}} \over {{a^2}}}\left( {{2 \over {\sqrt 2 }}} \right) = {{G{M^2}} \over {{a^2}}}\left( {{1 \over 2} + \sqrt 2 } \right)$$

This force will act as centripetal force.

Distance of particle from centre of circle is $${a \over {\sqrt 2 }}$$.

Here, $${F_{centripetal}} = {{M{v^2}} \over r} = {{M{v^2}} \over {{a \over {\sqrt 2 }}}} = {{\sqrt 2 M{v^2}} \over a}$$ ($$\because$$ $$r = {a \over {\sqrt 2 }}$$)

So, for rotation about the centre,

$${F_{centripetal}} = {F_{BO(net)}}$$

$$ \Rightarrow \sqrt 2 {{M{v^2}} \over a} = {{G{M^2}} \over {{a^2}}}\left( {{1 \over 2} + \sqrt 2 } \right)$$

$$ \Rightarrow {v^2} = {{GM} \over a}\left( {1 + {1 \over {2\sqrt 2 }}} \right) = {{GM} \over a}(1.35)$$

$$ \Rightarrow v = 1.16\sqrt {{{GM} \over a}} $$