Solution

The method of solution is identical to that in problem 12.1.6. In the rectangular contour, let's denote the sides as $ab = L$ and $aa' = x$. Faraday's law of electromagnetic induction in magnitude for the contour is:
$$\mathscr{E}_i = \frac{d\Phi}{dt}$$
where $\mathscr{E}_i$ is the induced electromotive force (EMF) created by the electric field $E$, and $\Phi$ is the magnetic flux through the contour.

From the geometry of the setup:
$$\mathscr{E}_i = E L$$
$$d\Phi = B L dx$$

Substituting these two formulas into the induction law, we get after canceling $L$:
$$E = B \frac{dx}{dt}$$

Here, the rate of change of distance $dx/dt$ is simply the propagation velocity of the wave in the medium, $v = c/\sqrt{\varepsilon}$. Therefore:
$$E = B \frac{c}{\sqrt{\varepsilon}} \implies B = \frac{E\sqrt{\varepsilon}}{c}$$
This is the magnetic induction in the SI system.

In the CGS (Gaussian) system, Faraday's law of electromagnetic induction contains an additional factor of $1/c$ on the right side:
$$EL = \frac{1}{c} \frac{d\Phi}{dt} = \frac{1}{c} B L \frac{dx}{dt}$$
Substituting the velocity $dx/dt = c/\sqrt{\varepsilon}$:
$$EL = \frac{1}{c} B L \frac{c}{\sqrt{\varepsilon}} = B L \frac{1}{\sqrt{\varepsilon}}$$
Canceling $L$, we find the magnetic induction in the CGS system:
$$E = B \frac{1}{\sqrt{\varepsilon}} \implies B = E\sqrt{\varepsilon}$$