The pulleys and strings shown in the figure are smooth and of negligible mass. For the system to remain in equilibrium, the angle $\mathrm{\theta }$ should be

Due to symmetry, the tension in both the strings is the same. Let it be T. The force on the mass $\sqrt{2}\mathrm{m} \mathrm{are} 2\mathrm{Tcos\theta }$ upwards and $\sqrt{2} \mathrm{mg}$ downwards. Since the mass is in equilibrium, $2\mathrm{Tcos} \mathrm{\theta }=\sqrt{2}\mathrm{mg}$. Further, since each mass m is also in equilibrium, 

$\mathrm{T}=\mathrm{mg}.$ Therefore, $2\mathrm{mg} \cos \mathrm{\theta }=\sqrt{2} \mathrm{mg}\Rightarrow \cos \mathrm{\theta }=1/\sqrt{2} \mathrm{or} \mathrm{\theta }={45}^0$