Answer
$0.14C^{\circ}$
Work Step by Step
There is more KE in the beginning; The KE lost goes toward heating the squash ball.
$$\frac{1}{2}m(v_i^2-v_f^2)=mc_{rubber}\Delta T$$
$$\Delta T=\frac{(v_i^2-v_f^2)}{2 c_{rubber}}=\frac{(22m/s)^2-(12m/s)^2}{2 (1200 J/(kg\cdot C^{\circ}))}=0.14C^{\circ}$$
