Answer
$E_{rms}=18.965\frac{N}{C}$
$B_{rms}=6.32\times 10^{-8}T$
Work Step by Step
As $I_{avg}=\frac{P_{avg}}{A}$
We plug in the known values to obtain:
$I_{avg}=\frac{75}{(2.5)^2}=0.95\frac{W}{m^2}$
Now the rms of the electric and magnetic field can be determined as
$E_{rms}=\sqrt{\frac{I_{avg}}{c\epsilon_{\circ}}}$
We plug in the known values to obtain:
$E_{rms}=\sqrt{\frac{0.95}{3\times 10^8\times 8.85\times 10^{-12}}}=18.965\frac{N}{C}$
Similarly $B_{rms}=\frac{E_{rms}}{c}$
We plug in the known values to obtain:
$B_{rms}=\frac{18.965}{3\times 10^8}=6.32\times 10^{-8}T$
