Good to see you solved the paradox yourself! Another way to find the same result is to compute the number of orbits which is possible to stack in a surface equal to the area of the system.
Let's first remind basic things on Landau levels. The hamiltonian of a particle moving in 2D $\{x,y\}\equiv\{r,\theta\}$ plan through a static magnetic field reads :
$$
\mathcal{H}=\frac{(\textbf{p}-q\mathbf{\mathcal{A}})^2}{2M}=\frac{\textbf{p}^2}{2M}+\frac{1}{8}M\,\omega^2_c\,r^2-\frac{\omega_c}{2}\,L_z
$$
where $\mathcal{A}=(-By/2,Bx/2,0)$ is the potential vector in the symmetric gauge, $L_z=xp_y-yp_x=-\mathrm{i}\hbar\,\partial_\theta$ the canonical angular momentum, and $\omega_c=qB/M$ the cyclotron pulsation.
Heisenberg inequalities provide here consistant physical constants of the problem (typical length $\ell$ and velocity $v_m$ of the movement) :
$$
M\,v_m\,\ell \sim \hbar \quad\text{with}\quad v_m=\ell\,\omega_c
$$
Thus, we find $\ell=\sqrt{\frac{\hbar}{M\omega_c}}$ which is often refered as the magnetic length.
One can then compute the spectrum of $\mathcal{H}$ :
$$
\mathcal{Sp(H)}=\left[ E_{n,m}=\left(n-m+1\right)\frac{\hbar\omega_c}{2};n \geq 0, m=-n,-n+2 \,...\,n-2,n\right]
$$
where $n$ is the quantum number associated to the $\frac{\textbf{p}^2}{2M}+\frac{1}{8}M\,\omega^2_c\,r^2$ part of $\mathcal{H}$, and $m$ is the magnetic quantum number. Note that for a movement in the entire plan $\mathbb{R}^2$, Laundau Levels degeneracy $D$ is infinite.
If we now restrain our discussion to the Lowest Landau Level (LLL) given for $n=m$, it is easy to check that the eigen wave function is something like :
$$
\Psi_{n=m}(r,\theta)\propto \left(r\,e^{\mathrm{i}\theta}\right)^m e^{-(r/2\ell)^2}
$$
The essential features of $\Psi_{n=m}$ is that its RMS width is nothing more than the magnetic length $\ell$ and that it is maximum around a radius $r_{\mathrm{max}}=\sqrt{2m+1}\ell$.
If we now consider that the movement of the particle is confined in a $R$ radius disk, the degeneracy $D$ of the LLL can be estimated by counting how many orbits one can put in the surface $S=\pi\,R^2$ of the system. Suppose that the system is large enough $S>>\ell^2$ to receive a large number $m>>1$ of orbitals, then according to the expression of $\Psi_{n=m}$, the condition to satisfy to be able to put these orbitals in $S$ is simply :
$$
R^2 > m\,2\ell^2
$$
Thus, the degeneracy is simply :
$$
D\sim\frac{R^2}{2\ell^2}=\frac{S}{2\pi\ell^2}=\frac{\Phi}{\Phi_0}
$$
$2\pi\ell^2$ can easily interpreted as the typical surface of an orbital.
