c
Length of the wire, \(l=3 \,c m=0.03 \,m\)

Current flowing in the wire, \(I=10 \,A\)

Magnetic field, \(B =0.27 \,T\)

Angle between the current and magnetic field, \(\theta=90^{\circ}\) ( Because magnetic field produced by a solenoid is along its axis and current carrying wire is kept perpendicular to the axis)

Magnetic force exerted on the wire is given as

\(F=B I \sin \theta\)

\(=0.27 \times 10 \times 0.03 \sin 90^{\circ}\)

\(=8.1 \times 10^{-2} \;N\)

Hence, the magnetic force on the wire is \(8.1 \times 10^{-2} \,N\). The direction of the force can be obtained from Fleming's left hand rule.