Please explain the conversion

Solution,
Average intensity of a beam is the average energy crossing per unit time per unit area about the point in a direction perpendicular to the direction of propagation of the beam. It is given as,
$$\begin{gathered} I_{av}=U_{av}c \\ where, \\ {U}_{av}=\frac{1}{2}{\epsilon }_o{E}_o^2 \\ {E}_o is the maximum value of electric field in the beam \\ c is the speed of light \\ Thus, \\ {I}_{av}=\frac{1}{2}{\epsilon }_o{E}_o^{2}c = \frac{1}{2}\times 8.85\times {10}^{-12}\times {36}^2\times 3\times {10}^8 = 1.72 \end{gathered}$$

Thus, Ans (3)
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