1. We need the unit of

$\sqrt{\frac{2I}{\varepsilon_0 c}}$

where:

• $I$ = intensity of electromagnetic wave
• $\varepsilon_0$ = permittivity of free space
• $c$ = speed of light

2. Use the electromagnetic wave intensity relation:

$I = \frac{1}{2}\varepsilon_0 c E_0^2$

Rearranging,

$E_0^2 = \frac{2I}{\varepsilon_0 c}$

So,

$\sqrt{\frac{2I}{\varepsilon_0 c}} = E_0$

Thus the given expression has the unit of electric field.

3. Unit of electric field:

$E = \frac{F}{q}$

Hence SI unit is

$\text{N C}^{-1}$

It is also equal to $\text{V m}^{-1}$, but among the given options only $\text{N C}^{-1}$ appears.

4. Check options:

• A: $\text{V m}$ → incorrect
• B: $\text{N C}^{-1}$ → correct
• C: $\text{N C}$ → incorrect
• D: $\text{N m}$ → incorrect

Therefore, the correct option is B.