If the average force of the cue on the ball was $200\ \mathrm{N}$, over what distance did this force act?

$d=0,25\ \mathrm{cm}$

The kinetic energy of the ball as it leaves the cue is

$$K=\frac{1}{2}mv^2$$

$$K=\frac{1}{2}(0.25\ \mathrm{kg})\left(2\ \mathrm{\displaystyle \frac{m}{s}}\right)^2$$

$$K=0.50\ \mathrm{J}$$

The work $(W)$ done by the cue gave the ball this kinetic energy, so

$$W=\Delta K$$

$$W=K_{\mathrm{f}}$$

$$Fd=K$$

$$d=\frac{K}{F}$$

$$d=\frac{0.50\ \mathrm{J}}{200\ \mathrm{N}}$$

$$d=0.0025\ \mathrm{m}$$

$$d=0,25\ \mathrm{cm}$$