Using the constants $\mu_0$ (or $\varepsilon_0$), $c$, $\hbar$, $e$ and $G$, it is possible to define two quantities with units of magnetic field : \begin{align} B_1 &= \sqrt{\frac{\mu_0 c^7}{\hbar G^2}} \equiv \sqrt{\frac{c^5}{\varepsilon_0 \hbar G^2}} \approx 8 \times 10^{53} \, \mathrm{T}, \tag{1} \\[12pt] B_2 &= \frac{c^3}{G e} \approx 3 \times 10^{54} \, \mathrm{T}. \tag{2} \end{align} Which one is really the Planck magnetic field?
While $B_2$ is simpler, I suspect it should be $B_1$, because it doesn't use the electric charge unit. $e$ is not exactly as universal as $\mu_0$. $B_1$ uses the Planck constant, so it's consistent to call it a Planck "unit", while $B_2$ doesn't use that constant. Also, because of the square root, $B_1$ is a bit more of the same shape as the Planck length : \begin{equation}\tag{3} L_{P} \equiv \sqrt{\frac{\hbar G}{c^3}}. \end{equation} The Planck units are presented on wikipedia: https://en.wikipedia.org/wiki/Planck_units but it doesn't tell anything about the magnetic field.
We could also argue that $B_1$ is the answer because we can find it by equating the magnetic field energy density with the Planck density (dropping all the dimensionless constants) : \begin{equation} \frac{B_1^2}{2 \mu_0} = \frac{M_P c^2}{L_P^3}. \end{equation} But then, we could also find $B_2$ by equating the Planck cyclotron angular frequency with the Planck energy : \begin{equation} \hbar \omega_{\text{cyclotron}} \equiv \hbar \, \frac{e B_2}{2 M_P} = M_P c^2. \end{equation} Both methods are arbitrary.
So what is the Planck magnetic strength?