For the work-energy theorem, the work done must be equal to the variation of kinetic energy of the ball:
[tex]W=\Delta K = \frac{1}{2} m (\Delta v)^2 [/tex]
where [tex]m=52 g=0.052 kg[/tex] is the mass of the ball, and where [tex]\Delta v[/tex] is the variation of velocity of the ball.
The initial velocity of the ball is +24 m/s, while the final velocity is -24 m/s (the negative sign is due to the fact the ball goes now in the opposite direction), so the variation of velocity is
[tex]\Delta v=24 m/s-(-24 m/s)=48 m/s[/tex]
Therefore, substituting all the numbers in the formula we find the work done:
[tex]W= \frac{1}{2}(0.052 kg)(48 m/s)^2=59.9 J [/tex]
