\(\mathrm{M}=\mathrm{NiA} \cos \theta\) i.e., \(\mathrm{M} \propto \cos \theta\)

When two magnetic fields are inclined at an angle of \(75^{\circ}\) the equilibrium will be at \(30^{\circ},\) so

\(\cos \theta=\cos \left(75^{\circ}-30^{\circ}\right)=\cos 45^{\circ}=\frac{1}{\sqrt{2}}\)

\(\frac{x}{\sqrt{2}}=\frac{15}{2} \therefore x \approx 11\)