Find amplitude of electric field in parallel beam of light of intensity  $\frac{6}{10\pi }\text{\hspace{0.17em}W}/{\text{m}}^{\text{2}}$. and take  $\frac{1}{4\pi {\epsilon }_0}=9\times {10}^9{\text{Nm}}^2/{\text{C}}^2$

Given
Intensity of light  $I=2.0\text{\hspace{0.17em}W}/{\text{m}}^{\text{2}}$
We know intensity of light  $I=\frac{1}{2}{\epsilon }_0{E}_0^{2}C$
Where ${\epsilon }_0=$ Permitivity= $8.85\times {10}^{-12}\text{\hspace{0.17em}F}/\text{m}$
      $C=$Speed of light= $3\times {10}^8\text{\hspace{0.17em}m}$
$\therefore I=\frac{1}{2}\times 8.85\times {10}^{-12}\times E_0^2\times 3\times {10}^8$

$\Rightarrow \frac{6}{10\pi }=\frac{1}{2}\times 8.85\times {10}^{-12}\times E_0^2\times 3\times {10}^8$

$\Rightarrow E_0^2=144$

$\Rightarrow E_0=12 V/m$