b
Magnetic field strength, \(B=0.25 \,T\)

Torque on the bar magnet, \(T=4.5 \times 10^{-2} \,J\)

Angle between the bar magnet and the external magnetic field, \(\theta=30^{\circ}\)

Torque is related to magnetic moment \((M)\) as:

\(T=M B \sin \theta\)

\(\therefore M=\frac{T}{B \sin \theta}\)

\(=\frac{4.5 \times 10^{-2}}{0.25 \times \sin 30^{\circ}}=0.36 \,J\,T ^{-1}\)

Hence, the magnetic moment of the magnet is \(0.36 \,J\,T ^{-1}\)