A thin bar magnet is bent in the form of a semicircle  and  its magnetic moment is M. If it is straightened and given a shape of an arc which subtends an angle ${120}^0$ at the center then its new magnetic moment will be 

When a bar magnet of moment $M_0$ is bent in the form of an arc its moment becomes $\frac{2M_0\sin \left({\displaystyle \frac{\theta }{2}}\right)}{\theta }$ where $\theta$ is the angle subtended at the center. here $\theta$ in the first case is 18$0^0$ and second case it is ${120}^0$

 $\frac{M\text{'}}{M}=\frac{sin\left({\displaystyle \frac{120}{2}}\right)}{\sin \left({\displaystyle \frac{180}{2}}\right)}\frac{180}{120}=\frac{\sqrt{3}}{2}\frac{3}{2}=\frac{3\sqrt{3}}{4}$