(a)
Hint: \(c=\frac{E_{0}}{B_{0}}\)
Step 1: Magnitude of magnetic field strength.

${\mathrm{B}}_0=\frac{{\mathrm{E}}_0}{\mathrm{c}}=\frac{120}{3\times {10}^8}=4\times {10}^{-7} \mathrm{T}=400 \mathrm{nT}$

Step 2: Find the angular frequency of the source.

$\omega =2\pi \nu =2\pi \times 50x106==3.14\times 108rad/s$

Step 3: Find the propagation constant.

$\mathrm{k}=\frac{\mathrm{\omega }}{\mathrm{c}}=\frac{3.14\times {10}^8}{3\times {10}^8}=1.05 \mathrm{rad}/\mathrm{m}$

Step 4: Find the wavelength.

$$$\mathrm{\lambda }=\frac{\mathrm{c}}{\mathrm{\nu }}=\frac{3\times {10}^8}{50\times {10}^6}$= 6.0 m

(b)
Hint: Electric field vector and the magnetic field vector are mutually perpendicular.
Step 1: Identify the direction of electric field vector and magnetic field vector.
If the wave is propagating in the positive x-direction, the electric field vector will be in the positive y-direction and the magnetic field vector will be in the positive z-direction.
Step 2: Find the Equation of electric field vector.

Equation of electric field vector is given by:

$$$\overrightarrow{\mathrm{E}}={\mathrm{E}}_0\sin (\mathrm{kx}-\mathrm{\omega t})\hat{\mathrm{j}}=120\sin \left[1.05\mathrm{x}-3.14\times {10}^8\mathrm{t}\right]\hat{\mathrm{j}}$


$$
Step 3: Find the Equation of magnetic field vector.
\(\vec{B}=B_{0}~sin\left ( kx-\omega t \right )\hat{k}\)
\(\vec{B}=\left ( 4 \right )\times 10^{-7}~sin\left [ 1.05x-3.14\times 10^{8}t \right ]\hat{k}\)