#### Answer

$F_{max} = 960~N$

#### Work Step by Step

If the ball hits the wall at 32 m/s and rebounds with the same speed, then the change in velocity is 64 m/s. We can find the impulse $J$ that the wall exerted on the ball. Note that the impulse is equal to the change in momentum.
$J=\Delta p$
$J = m~\Delta v$
$J = (0.060~kg)(64~m/s)$
$J = 3.84~N~s$
The impulse is equal to the area under the force versus time graph.
$area = J$
$\frac{1}{2}(F_{max})(2~ms)+(F_{max})(2~ms)+\frac{1}{2}(F_{max})(2~ms) = J$
$(2)(F_{max})(2~ms) = J$
$F_{max} = \frac{J}{(2)(2~ms)}$
$F_{max} = \frac{3.84~N~s}{(2)(2\times 10^{-3}~s)}$
$F_{max} = 960~N$