Answer
$F_{max} = 0.96~N$
Work Step by Step
We can find the impulse $J$ that the wall exerted on the ball as:
$p_0+J=p_f$
$J = p_f-p_0$
$J = m~(v_f-v_0)$
$J = (0.060~kg)[(-32~m/s)-(32~m/s)]$
$J = -3.84~N~s$
The magnitude of the impulse is equal to the area under the force versus time graph;
$area = 3.84~N~s$
$\frac{1}{2}F_{max}(2~s)+F_{max}(2~s)+\frac{1}{2}F_{max}(2~s) = 3.84~N~s$
$2~F_{max}(2~s) = 3.84~N~s$
$F_{max} = \frac{3.84~N~s}{(2)(2~s)}$
$F_{max} = 0.96~N$
