(A) Comparing the given equation with the standard form $B_y = B_0 \sin(kx + \omega t)$:
Here,$k = 0.5 \times 10^3 \, rad/m$ and $\omega = 1.5 \times 10^{11} \, rad/s$.
Wavelength $\lambda = \frac{2\pi}{k} = \frac{2 \times 3.14}{0.5 \times 10^3} \approx 1.26 \, m$.
Frequency $\nu = \frac{\omega}{2\pi} = \frac{1.5 \times 10^{11}}{2 \times 3.14} \approx 2.39 \times 10^{10} \, Hz = 23.9 \, GHz$.
$(b)$ The amplitude of the electric field is $E_0 = B_0 c = (2 \times 10^{-7} \, T) \times (3 \times 10^8 \, m/s) = 60 \, V/m$.
Since the wave propagates in the negative $x$-direction and the magnetic field is along the $y$-axis,the electric field must be along the $z$-axis. Thus,the expression is:
$E_z = 60 \sin (0.5 \times 10^3 x + 1.5 \times 10^{11} t) \, V/m$.