Let $a = 6$. Let $X$ be the slotted section and $Y$ be the slot.
Change coordinates so that $X$ and $Y$ are living on the plane $\mathcal{P} = \{ (x,y,z) : z = a \}$ and the reference point for computing the solid angle is located at the origin. The geometric setup is basically the same as in this answer.
Please read the method 3 in that answer first, in particular the Lemma there. We are going to need it later.
For any geometric shape $Z \subset \mathcal{P}$, let $\Omega_Z$ be the solid angle
subtended by $Z$. Consider the projection $\eta : \mathcal{P} \to S^2$ as in above answer.
$$\mathcal{P} \ni (x,y,a) \quad\mapsto\quad \frac{1}{\sqrt{x^2+y^2+a^2}}(x,y,a) \in \mathcal{S}^2$$
It is clear $\Omega_Z$ is simply the area of $\eta(Z)$ on $S^2$.
The key observation is under $\eta$, straight lines on $\mathcal{P}$ get mapped to geodesics on $S^2$.
So the boundary of $Y$ get mapped to a polygon of 16 sides on $S^2$ and $12$ of them are geodesics.
We will use Gauss-Bonnet theorem to evaluate $\Omega_X$ and $\Omega_Y$. In particular, we will use following version of the theorem on $S^2$:
Let $\gamma$ be a simply closed piece-wise smooth curve on $S^2$. Let $\theta_i$ be the external angles at the discontinuity of $\gamma'$. The area enclosed by $\gamma$ is given by the formula:
$$\verb/Area/ = 2\pi - \left( \int_\gamma k_g ds + \sum_i \theta_i \right)$$
where $k_g$ is the geodesic curvature for $\gamma$.
If $D_\rho$ is a disk on $\mathcal{P}$ centered at $O = (0,0,a)$ with radius $\rho$, one can show that
$$\int_{\eta(\partial D_\rho)} k_g ds = 2\pi\frac{1}{\sqrt{1+(\rho/a)^2}}$$
It is clear $X \cup Y = D_R$ with $R = 20$, we find
$$
\Omega_X + \Omega_Y =
\Omega_{D_R} = 2\pi \left( 1 - \frac{a}{\sqrt{a^2+R^2}} \right) = 2\pi\left(1 - \frac{3}{\sqrt{109}}\right)
$$
What's remain is to compute $\Omega_Y$. Since the inner radius $r = 5$ is the same as the width of the beam, the four curved sections of $\partial Y$ contribute one-third of that of a full circle. Since $k_g = 0$ on geodesics, we find
$$\int_{\partial Y} k_g ds = \frac{2\pi}{3}\frac{a}{\sqrt{a^2+r^2}} = 2\pi\left(\frac{2}{\sqrt{61}}\right)$$
To compute the $16$ external angles, let us walk around $\partial Y$ counter-clockwisely.
Eight of the external angles look like what happen at vertex $A$ (please refer to the figure below).
The vertex $A$ is a distance $\rho_A = 5$ from $O$. The incoming curve is perpendicular to the radial direction while the outgoing straight edge is making an angle $\frac{\pi}{6}$.
Using the Lemma in method 3 of the aforementioned answer, the external angle
at $A$ is equal to
$$
\theta_A
= \tan^{-1}\left(\sqrt{1 + (\rho_A/a)^2}\tan\frac{\pi}{3}\right) - \frac{\pi}{2}
= \tan^{-1}\left(\frac{\sqrt{183}}{18}\right) - \frac{\pi}{2}
$$
The remaining eight external angles look like what happen at vertex $B$.
The vertex $B$ is at a distance $\rho_B = \sqrt{15^2 + (5/2)^2} = \frac52\sqrt{37}$ from $O$. The incoming edge is making an angle $\tan^{-1}\frac16$ to the radial direction while the outgoing edge is making an angle $\tan^{-1}6$ to the radial direction. The external angle at $B$ is equal to
$$\begin{align}
\theta_B &=
\left( \pi - \tan^{-1}\left(6\sqrt{1+ (\rho_B/a)^2}\right) \right)
- \tan^{-1}\left(\frac16\sqrt{1+ (\rho_B/a)^2}\right)\\
&= \pi - \tan^{-1}\left(\frac{\sqrt{1069}}{2}\right) - \tan^{-1}\left(\frac{\sqrt{1069}}{72}\right)
= \tan^{-1}\left(\frac{2\sqrt{1069}}{25}\right)
\end{align}
$$
Combine all these, we find
$$\begin{align}
\Omega_Y & = 2\pi\left(1 - \frac{2}{\sqrt{61}}\right)
- 8\left(
\tan^{-1}\left(\frac{\sqrt{183}}{18}\right) +
\tan^{-1}\left(\frac{2\sqrt{1069}}{25}\right)
- \frac{\pi}{2}
\right)\\
\implies \Omega_X & =
8\left(
\tan^{-1}\left(\frac{2\sqrt{1069}}{25}\right)
+ \tan^{-1}\left(\frac{\sqrt{183}}{18}\right)
\right)
- 2\pi\left(
2 + \frac{3}{\sqrt{109}} - \frac{2}{\sqrt{61}}
\right)\\
& \approx 2.038049336500276
\end{align}
$$
