Answer
The wavelength of the AM wave is larger by a factor of $1.00\times10^2$.
Work Step by Step
See problem 34. The two frequencies are 980 kHz (AM) and 98.1 MHz (FM).
Use equation 22–4, $c=\lambda f$.
The lower frequency (AM) has the longer wavelength.
$$\lambda_{AM}=\frac{c}{f}=\frac{3.00\times10^8m/s }{980\times10^3Hz}$$
$$\lambda_{FM}=\frac{c}{f}=\frac{3.00\times10^8m/s }{98.1\times10^6Hz}$$
Divide to find the ratio.
$$\frac{\lambda_{AM}}{\lambda_{FM}}=\frac{98.1\times10^6Hz }{980\times10^3Hz }=1.00\times10^2$$