Answer
$E_0 = 980~V/m$
$B_0 = 3.3\times 10^{-6}~T$
Work Step by Step
We can find the intensity of the light wave:
$I = \frac{P}{A}$
$I = \frac{P}{\pi r^2}$
$I = \frac{1.0\times 10^{-3}~W}{(\pi) (0.50\times 10^{-3}~m)^2}$
$I = 1273~W/m^2$
We can find the electric field amplitude:
$I = \frac{1}{2}~E_0^2~\epsilon_0~c$
$E_0^2 = \frac{2~I}{\epsilon_0~c}$
$E_0 = \sqrt{\frac{2~I}{\epsilon_0~c}}$
$E_0 = \sqrt{\frac{(2)(1273~W/m^2)}{(8.854\times 10^{-12}~F/m)~(3.0\times 10^8~m/s)}}$
$E_0 = 980~V/m$
We can find the magnetic field amplitude:
$B_0 = \frac{E_0}{c}$
$B_0 = \frac{980~V/m}{3.0\times 10^8~m/s}$
$B_0 = 3.3\times 10^{-6}~T$
