(a)
Hint: Electromagnetic wave propagates perpendicular to the electric field.
Step: Find the direction of electromagnetic wave propagation.
From the given equation of the electric field vector, we can see that the electromagnetic wave is propagating along the negative y-direction.

(b)
Hint: \(\lambda =\frac{2\pi}{k}\)
Step 1: Compare the given equation with standard equation.

$\overrightarrow{\mathrm{E}}=\left\{\left(3.1 \mathrm{N}/\mathrm{C}\right) \cos \left[(1.8 \mathrm{rad}/\mathrm{m})\mathrm{y}+(5.4\times {10}^8\mathrm{rad}/\mathrm{s})\mathrm{t}\right]\right\}$\(\widehat i\)
Comparing it with:
$\overrightarrow{\mathrm{E}}={\mathrm{E}}_0 \sin (\mathrm{ky}+\mathrm{\omega t})\hat{\mathrm{i}}$

$\mathrm{Amplitude}of\mathrm{electric} \mathrm{field} {\mathrm{E}}_0= 3.1 \mathrm{N}/\mathrm{C}$
$\mathrm{Angular} \mathrm{frequency} \mathrm{\omega }=5 \times {10}^8 \mathrm{rad}/\mathrm{s }$
$\mathrm{Wave}\mathrm{number} \mathrm{k}= 1.8 \mathrm{rad}/\mathrm{m}$
$$
Step 2: Find the wavelength.
\(\lambda =\frac{2\pi}{k}\)=\(\frac{2\pi}{1.8}=3.490\) m

(c)
Hint: $\mathrm{\nu }=\frac{\mathrm{\omega }}{2\mathrm{\pi }}$
$\mathrm{}$Step: Find the frequency
$\mathrm{}$$\mathrm{\nu }=\frac{\mathrm{\omega }}{2\mathrm{\pi }}=\frac{5.4\times {10}^8}{2\mathrm{\pi }}=8.6\times {10}^7 \mathrm{Hz}$

$$ $$
(d)
Hint: ${\mathrm{B}}_0=\frac{{\mathrm{E}}_0}{\mathrm{c}}$
Step: Find the amplitude of the magnetic field.

${\mathrm{B}}_0=\frac{{\mathrm{E}}_0}{\mathrm{c}}$
${\mathrm{B}}_0=\frac{3.1}{3\times {10}^8}=1.03\times {10}^{-7} \mathrm{T}$

(e)
Hint: Electromagnetic wave propagates perpendicular to the electric and magnetic field.
Step: Find the equation of magnetic field.
As the wave is traveling along negative y-direction and the electric field is directed along the +x-direction, the magnetic field should direct towards +z-direction. Hence, the general equation for the magnetic field vector is:
$\overrightarrow{\mathrm{B}}={\mathrm{B}}_0\cos (\mathrm{ky}+\mathrm{\omega t})\hat{\mathrm{k}}$
$=\left\{\right(1.03\times {10}^{-7}\mathrm{T})\cos \left[(1.8 \mathrm{rad}/\mathrm{m})\mathrm{y}+(5.4\times {10}^6 \mathrm{rad}/\mathrm{s})\mathrm{t}\right]\}\hat{\mathrm{k}}$