You should remember that the impulse of collision is equal to the change in momentum.
Since the problem provides the mass of the ball and the speed of the ball before and after the collision, you may evaluate the momentum before collision `M_1` and momentum after collision,`M_2` , such that:
`M_1 = 60*10^-3*32 = 0.06*32 = 1.92 (Kg*m)/s`
`M_2 = 60*10^-3*(-32) = -1.92 (Kg*m)/s`
`Delta M = M_1 - M_2 = 1.92 - (-1.92) = 3.84 (Kg*m)/s`
Since the problem provides the graph of force, you may evaluate the impulse of collision as area under the graph, such that:
Impulse = `(6+2)/2*10^(-3)*F_max`
Notice that 6 and 2 represents the lengths of the parallel sides of trapezoid formed by the graph of force with x axis.
`Impulse = Delta M => (6+2)/2*10^(-3)*F_max = 3.84`
`F_max = 3.84/(4*10^(-3)) = 960 N`
Hence, evaluating the magnitude of the contact force during collision yields `F_max = 960 N` .