Answer
$B_{rms} = 2.5\times 10^{-16}~T$
Work Step by Step
We can find the maximum value of the electric field vector:
$I = \frac{E_m^2}{2c\mu_0}$
$E_m^2 = 2c~\mu_0~I$
$E_m = \sqrt{2c~\mu_0~I}$
$E_m = \sqrt{(2)(3.0\times 10^8~m/s)~(4\pi\times 10^{-7}~H/m)~(1.5\times 10^{-17}~W/m^2)}$
$E_m = 1.06\times 10^{-7}~V/m$
We can find $B_{rms}$:
$B_{rms} = \frac{E_{rms}}{c}$
$B_{rms} = \frac{E_m}{\sqrt{2}~c}$
$B_{rms} = \frac{1.06\times 10^{-7}~V/m}{(\sqrt{2})~(3.0\times 10^8~m/s)}$
$B_{rms} = 2.5\times 10^{-16}~T$
