Answer
$E_{max}=61.4V/m$
Work Step by Step
We know that
$E_{max}^2=\frac{2I_{avg}}{c\epsilon_{\circ}}$
$\implies E_{max}=\sqrt{\frac{2I_{avg}}{c\epsilon_{\circ}}}$
We plug in the known values to obtain:
$E_{max}=\sqrt{\frac{2(5.002/m^2)}{(3\times 10^8m/s)(8.85\times 10^{-12}C^2/Nm^2)}}$
$E_{max}=61.4V/m$