(b)For equilibrium of the system torques on $M_1$ and $M_2$ due to $B_H$ must counter balance each other 

i.e. ${M_1} \times {B_H} = {M_2} \times {B_H}$. If $\theta$ is the angle between $M_1$ and $B_H$ will be $(90 – \theta )$; 

so ${M_1}{B_H}\sin \theta = {M_2}{B_H}\sin (90 – \theta )$

$ \Rightarrow \tan \theta = \frac{{{M_2}}}{{{M_1}}} = \frac{M}{{3M}} = \frac{1}{3} \Rightarrow \theta = {\tan ^{ – 1}}\left( {\frac{1}{3}} \right)$