Answer
$E_0 = 0.092~V/m$
Since $~~0.092~V/m \lt 0.10~V/m,~~$ it seems that the manufacturer's claims are true.
Work Step by Step
We can find the intensity of the signal:
$I = \frac{P}{4\pi~d^2}$
$I = \frac{0.250~W}{(4\pi)~(42~m)^2}$
$I = 1.1278\times 10^{-5}~W/m^2$
We can find the electric field amplitude:
$I = \frac{1}{2}~E_0^2~\epsilon_0~c$
$E_0^2 = \frac{2~I}{\epsilon_0~c}$
$E_0 = \sqrt{\frac{2~I}{\epsilon_0~c}}$
$E_0 = \sqrt{\frac{(2)(1.1278\times 10^{-5}~W/m^2)}{(8.854\times 10^{-12}~F/m)~(3.0\times 10^8~m/s)}}$
$E_0 = 0.092~V/m$
The door fails to respond when the electric field amplitude decreases to $0.092~V/m$
Since $~~0.092~V/m \lt 0.10~V/m,~~$ it seems that the manufacturer's claims are true.
