Answer
0.499 N.
Work Step by Step
As the loop is pulled from the field, a motional emf occurs, given by equation 21–3,
$$\epsilon = B\mathcal{l}v = (0.550T)(0.350m)(3.10m/s)=0.59675V$$
The induced current is clockwise, to make more magnetic field lines into the page, replacing those that are lost as the loop withdraws from the field.
$$I=\frac{\epsilon}{R}=\frac{0.59675V}{0.230\Omega}=2.59457 A$$
By the right hand rule, there is a net magnetic force to the left, $I\mathcal{l}B$. The external force to the right is equal and opposite.
$$F= I\mathcal{l}B =(2.59457A)(0.350m)(0.550T)=0.499N$$
