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**Hint:**In this question the concept of relative velocity is applied. Relative velocity is defined between two objects in motion. If the two objects are moving in opposite directions the relative velocity is obtained by adding the both velocities and if the two objects are moving in the same direction then the relative velocity is calculated by subtracting the velocity of one object from the other.

**Complete step by step answer:**

Let us consider the figure below shows the velocity and relative velocity.

Figure-(1)

In above figure the component of velocity of Shyam is shown in vertical as well as horizontal direction.

Velocity of Shyam is at an angle so the component will be taken.

The velocity of Shyam can be expressed as,

${V_s} = 6\sin 30^\circ i + 6\cos 30^\circ j$

Now we will simplify above expression.

$ \Rightarrow {V_s} = 6\left( {\dfrac{1}{2}} \right)i + 6\left( {\dfrac{{\sqrt 3 }}{2}} \right)j$

Further solve, we will get,

$ \Rightarrow {V_s} = 3i + 3\sqrt 3 j$

As we know that the relative velocity can be written as,

${V_{rs}} = {V_r} - {V_s}$

Here, ${v_{rs}}$ is the relative velocity of Ram with respect to Shyam.

Now we will substitute the values in the above equation.

$ \Rightarrow {V_{rs}} = 6i - 3i - 3\sqrt 3 j$

Further solve, we will get,

$ \Rightarrow {V_{rs}} = 3i - 3\sqrt 3 j$

Now we will calculate the magnitude of relative velocity,

$ \Rightarrow \left| {{V_{rs}}} \right| = \sqrt {{3^2} + {{\left( {3\sqrt 3 } \right)}^2}} $

Further simplify the calculation and we get,

$\therefore \left| {{V_{rs}}} \right| = 6\;m/s$

**Therefore, the correct option is B.**

**Note:**In this question the relative velocity will be calculated by breaking the velocity of Shyam in two components that are horizontal and vertical. Vector calculation is to be used instead of simple arithmetic operations. The sign of components of velocity should be used properly otherwise it will alter the final result.

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