The electric field in an electromagnetic wave is given by E = 56 ....

Given, $E = 56 . 5sin\omega\left(\dfrac{t - x}{c}\right)N{C}^{- 1}$

Comparing it with the general equation of electromagnetic wave,

$E = {E}_{0}\sin \left(\omegat - kx\right)$

We have, ${E}_{0} = 56 . 5N{C}^{- 1}$

So, the intensity will be $I = \dfrac{1}{2}{ε}_{0}{E}_{0}^{2}c$

$= \dfrac{1}{2} \times 8 . 85 \times {10}^{- 12} \times {\left(56 . 5\right)}^{2} \times 3 \times {10}^{8}$

$= 4 . 24W{m}^{- 2}$