Answer
The cannonball has greater kinetic energy.
Work Step by Step
Before the cannonball was fired, there is no motion, so $\sum p_0=0$
Take the cannonball's direction to be positive. According to the conservation of linear momentum, $$m_{cannon}(-v_{cannon})+m_{ball}v_{ball}=0$$ $$\frac{m_{cannon}}{m_{ball}}=\frac{v_{ball}}{v_{cannon}}(1)$$
We have $$\frac{KE_{cannon}}{KE_{ball}}=\frac{\frac{1}{2}m_{cannon}v_{cannon}^2}{\frac{1}{2}m_{ball}v_{ball}^2}=\frac{m_{cannon}v_{cannon}^2}{m_{ball}v_{ball}^2}$$
Plug (1) here: $$\frac{KE_{cannon}}{KE_{ball}}=\frac{v_{ball}}{v_{cannon}}\times\frac{v_{cannon}^2}{v_{ball}^2}=\frac{v_{cannon}}{v_{ball}}=\frac{m_{ball}}{m_{cannon}}\lt1$$ $$KE_{cannon}\lt KE_{ball} $$
