A magnetic needle suspended parallel to a magnetic field requires $\sqrt{3}J$ of work to turn it through ${60}^0.$ The torque required to maintain the needle in this position will be

Given, Work = $\sqrt{3}J$

Angle = ${60}^0$

We know, 

Work done in turning a magnet of magnetic moment 'M' by an angle $\mathrm{\theta }$ is:

$\mathrm{w}=\mathrm{MB}(\cos 0°-\mathrm{cos\theta })$ ----(1)

$\mathrm{\tau }=\mathrm{MBsin\theta }$ ----(2)

Dividing equations$\frac{\left(1\right)}{\left(2\right)}$:

$\frac{\mathrm{w}}{\mathrm{\tau }}=\frac{1-\mathrm{cos\theta }}{\mathrm{sin\theta }}$

$\frac{\sqrt{3}}{\mathrm{\tau }}=\frac{1-\mathrm{cos\theta }}{\mathrm{sin\theta }}$

$\frac{\sqrt{3}}{\tau }=\frac{1-(1/2)}{\sqrt{3}/2}$

$\frac{\sqrt{3}}{\tau }=\frac{1}{\sqrt{3}}$

$\mathbf{\tau }\mathbf{=}\mathbf{3}\mathbf{ }\mathbf{Nm}$