A magnetic needle lying parallel to a magnetic field requires W units of work to turn it through 60⁰.  The magnitude of torque needed to maintain the needle in this position will be

Work done to rotate the needle=Potential energy of needle=W

If M=Magnetic dipole moment of the needle and B=magnetic field then:

$W=MB(\cos {\theta }_1-\cos {\theta }_2) (where {\theta }_1=0^0 and {\theta }_2={60}^0)$$\Rightarrow W=MB(1-1/2)= \frac{ MB }{2} or MB=2W\to \left(1\right)$

So restoring torque acting on the needle is given by:

$\tau =M\times B=MB\sin {60}^0 (In magnitude)$

$\Rightarrow \tau =\frac{\sqrt{3}}{2}MB$

$\Rightarrow \tau =\frac{\sqrt{3}}{2}2W=\sqrt{3} W.$