For an electromagnetic wave in free space, the magnitudes of the electric and magnetic fields are related by the speed of light, $$c \approx 3 \times 10^8 \, \mathrm{m/s},$$ through the equation

$$ E = cB. $$

Here's how to determine the magnetic field:

We are given the electric field magnitude: $$E=9.3 \, \mathrm{V/m}.$$

Use the relationship to find the magnetic field magnitude:

$$ B = \frac{E}{c} = \frac{9.3}{3 \times 10^8} \, \mathrm{T}. $$

Calculating the value gives:

$$ B \approx 3.1 \times 10^{-8} \, \mathrm{T}. $$

Since the electric field is along the $$y$$ direction and the wave propagates along the $$+x$$ direction, by the right-hand rule, the magnetic field is oriented along the $$z$$ direction.

Thus, the magnetic field vector is

$$ \mathrm{B}_z = 3.1 \times 10^{-8} \, \mathrm{T}, $$

which corresponds to Option C.