The dimensional formula of $\frac{1}{2}{\epsilon }_0{E}^2$  (${\epsilon }_0$: permittivity of free space and  $E:$ Intensity of electric field) is

Given
Energy density of electric field  $=\frac{1}{2}{\epsilon }_0{E}^2$
Here ${\epsilon }_0=$ Permittivity of free space
$E=$Intensity of electric field
$\Rightarrow \left[\frac{1}{2}{\epsilon }_0{E}^2\right]=\left[\frac{\text{Energy}}{\text{Volume}}\right]=\frac{M{L}^2{T}^{-2}}{L^3}=M{L}^{-1}{T}^{-2}$
$\therefore \text{Dimension\hspace{0.17em}formula\hspace{0.17em}of\hspace{0.17em}}\frac{1}{2}{\epsilon }_0{E}^2=M{L}^{-1}{T}^{-2}$
Therefore, the correct answer is (C).