Let's first solve for the launch velocity. If the landing is at the same level as the cannon, the range of a projectile (assuming no air drag etc) is
$$
d = \frac{v^2}{g} \sin{2 \theta}
$$
where $v$ is the launch velocity, $g$ is the acceleration of gravity and $\theta$ is the launch angle. Solving for $v$, we obtain
$$
v_{\text{launch}} = \sqrt{ \frac{dg}{\sin{2\theta}} }
$$
Now let's look a the the barrel. It's essentially a linear path that the projectile goes through before launching into the air. The initial velocity is zero, so the following equation ca be written for the velocity of the projectile within the barrel:
$$
v(t) = at \qquad \Rightarrow \qquad t = \frac{v_{\text{launch}}}{a}
$$
where $a$ is a (constant) acceleration and $t$ is time. On the other hand, integrating this equation, we can obtain the equation for the distance (as a function of time) inside the barrel:
$$
s(t) = \frac{1}{2}at^2
$$
plugging in the previous result for time, we obtain
$$
s(t) = \frac{1}{2} a \left( \frac{v_{\text{launch}}}{a}\right)^2
= \frac{ v^2_{\text{launch}} }{2a}
$$
from which we can solve for the acceleration $a$:
$$
a = \frac{ v^2_{\text{launch}} }{2s}
$$
Here, $s$ is the length of the barrel. Now we can plug in the result for the launch velocity:
$$
a = \frac{ \left( \frac{dg}{\sin{2\theta}} \right) }{2s} = \frac{dg}{2s \, \sin{2\theta}}
$$
the "G" acceleration is specifically acceleration divided by $g$, so the "G" acceleration as a formula is
$$
\frac{a}{g} = \frac{d}{2s\, \sin{2 \theta}} = \frac{ \left(6.096~\text{m} \right)}{2 \left(0.203~\text{m} \right) \sin{90^{\circ}}} \approx 15~\text{g's}
$$
