Answer
$100\;\rm m/s$
Work Step by Step
We can use the conservation of momentum principle if we assumed that there is no friction and that the spring is an ideal one that obeys Hooke's law.
And to make it easy, let's assume that the ball is fired horizontally.
Thus,
$$p_{ix}=p_{fx}$$
where $p_{ix}$ is zero since the system was at rest just before the collision (firing the balls).
$$0=m_bv_b+m_cv_c $$
Solving for $v_b$;
$$v_b=\dfrac{-m_cv_c}{m_b}\tag 1$$
where $b$ is for the cannon's balls and $c$ is for the cannon itself.
Now we can assume that the energy is conserved as we did above since there is no friction.
The system here is the cannon+the spring.
$$E_i=E_f$$
where $E_i$ is the energy of the system just after firing the ball while $E_f$ is when the spring is compressed at a distance of 50 cm and the cannon stops.
$$U_{is}+K_i=U_{fs}+K_f$$
$$\frac{1}{2}kx_i^2+ \frac{1}{2}m_cv_c^2=\frac{1}{2}kx_f^2 +\frac{1}{2}m_cv_{fc}^2$$
As we mentioned above, the final velocity of the cannon is zero. The initially compressed distance of the spring was zero.
$$0+ \color{red}{\bf\not}\frac{1}{2}m_cv_c^2=\color{red}{\bf\not}\frac{1}{2}kx_f^2 +0$$
Solving for $v_c$;
$$v_c=\sqrt{\dfrac{kx_f^2}{m_c}}$$
Plugging the known;
$$v_c=\sqrt{\dfrac{20000 \times0.5^2}{200}}=\pm\bf 5\;\rm m/s$$
If we assumed that the ball is fired forward, the cannon is moved backward.
Thus, its velocity has to have a negative sign,
$$v_c=- \bf 5\;\rm m/s$$
Plugging into (1) and plug the other known;
$$v_b=\dfrac{-200 (-5)}{10} $$
$$v_b=\color{red}{ \bf 100}\;\rm m/s$$
