Answer
$43.4\frac{V}{m}$
Work Step by Step
We know that
$I_{max}=c\epsilon_{\circ}E_{max}^2$
This simplifies to:
$E_{max}=\sqrt{\frac{I_{max}}{c\epsilon_{\circ}}}$
We plug in the known values to obtain:
$E_{max}=\sqrt{\frac{5.00}{(3.00\times 10^8)(8.85\times 10^{-12})}}=43.4\frac{V}{m}$
