#### Answer

(a) $t = 180~s$
(b) The signal is detected at a wavelength of 488 nm

#### Work Step by Step

(a) We can find the time it takes the signal to reach the planet:
$t = \frac{d}{v}$
$t = \frac{54\times 10^6~km}{3.0\times 10^5~km/s}$
$t = 180~s$
(b) We can find the wavelength that the signal is detected:
$\lambda' = \lambda_0~\frac{\sqrt{1-v/c}}{\sqrt{1+v/c}}$
$\lambda' = (540~nm)~\frac{\sqrt{1-0.1c/c}}{\sqrt{1+0.1c/c}}$
$\lambda' = (540~nm)~\frac{\sqrt{0.9}}{\sqrt{1.1}}$
$\lambda' = 488~nm$
The signal is detected at a wavelength of 488 nm.