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.