The unit of $$\sqrt{\frac{2I}{\epsilon_0 c}}$$ is : (I = intensity of an electromagnetic wave, c : speed of light)

The intensity of a plane electromagnetic wave is related to the peak electric-field amplitude $$E_0$$ by

$$I = \frac{1}{2}\,\epsilon_0\,c\,E_0^{2}$$ $$-(1)$$

Solving $$-(1)$$ for $$E_0$$ gives

$$E_0 = \sqrt{\frac{2I}{\epsilon_0 c}}$$ $$-(2)$$

Thus the expression whose unit we must find is simply the unit of the electric field $$E_0$$.

Unit analysis:

Intensity  $$I$$ : power per area   $$\Rightarrow$$  $$\text{W m}^{-2} = \left(\text{J s}^{-1}\right)\text{m}^{-2} = \text{kg s}^{-3}$$

Permittivity  $$\epsilon_0$$ : $$\text{C}^{2}\,\text{N}^{-1}\,\text{m}^{-2}$$ Force  $$\text{N}= \text{kg m s}^{-2}$$ $$\therefore \epsilon_0 = \frac{\text{C}^{2}}{\text{kg m s}^{-2}\,\text{m}^{2}} = \frac{\text{C}^{2}\,\text{s}^{2}}{\text{kg m}^{3}}$$

Speed of light  $$c$$ : $$\text{m s}^{-1}$$

Compute the unit of $$\dfrac{2I}{\epsilon_0 c}$$ :

$$ \dfrac{\text{kg s}^{-3}} {\left(\dfrac{\text{C}^{2}\,\text{s}^{2}}{\text{kg m}^{3}}\right)\;(\text{m s}^{-1})} = \dfrac{\text{kg s}^{-3} \;\text{kg m}^{3}} {\text{C}^{2}\,\text{s}^{2}\;\text{m}} = \dfrac{\text{kg}^{2}\,\text{m}^{2}\,\text{s}^{-4}}{\text{C}^{2}} $$

Taking the square root (see $$-(2)$$) gives the unit of $$E_0$$:

$$ \sqrt{\dfrac{\text{kg}^{2}\,\text{m}^{2}\,\text{s}^{-4}}{\text{C}^{2}}} = \dfrac{\text{kg m s}^{-2}}{\text{C}} = \dfrac{\text{N}}{\text{C}} $$

Hence $$\sqrt{\dfrac{2I}{\epsilon_0 c}}$$ has the SI unit $$\mathbf{N\,C^{-1}}$$, which is also equal to $$\mathbf{V\,m^{-1}}$$.

Answer : Option D  $$NC^{-1}$$