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**Hint:**Find the relation between electric field and magnetic field. They are related to each other by c i.e. speed of light, Write the equation for c in terms of wavelength. Then, get the expression for c in terms of angular frequency and wave number. Finally, substitute that value in the relation of electric field and magnetic field.

**Complete answer:**

Given: Equations of electromagnetic waves: $E= { E }_{ 0 } \sin { \left( kx-\omega t \right) }$ and $B={ B }_{ 0 }\sin { \left( kx-\omega t \right) }$

The relation between Electric field ${ E }_{ 0 }$ and Magnetic field ${ B }_{ 0 }$ is given by,

${ E }_{ 0 }=c{ B }_{ 0 }$ …(1)

We know, $\nu =\dfrac { c }{ \lambda }$

Rearranging the above equation we get,

$c=\nu \lambda$ …(2)

We also know, $\nu =\dfrac { \omega}{ 2\pi }$

Thus, substituting this value in the equation. (2) we get,

$c=\dfrac { \omega}{ 2\pi }\lambda$

Now, substituting $\cfrac { 2\pi }{ \lambda } =k$ we get,

$c=\dfrac { \omega }{ k }$ …(2)

Then, substituting the equation. (3) in equation. (1) we get,

${ E }_{ 0 }=\dfrac { \omega }{ k } { B }_{ 0 }$

Rearranging the above equation we get,

${ E }_{ 0 }k={ B }_{ 0 }\omega$

**So, the correct answer is “Option A”.**

**Note:**

The direction of electromagnetic waves is found by vector cross product of the electric field and magnetic field. Electromagnetic radiations also show dual nature i.e. have the properties of a wave as well as the properties of a particle.

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