Answer
The wavelength is $6.6 \times 10^{-30} m$, which is about $3 \times 10^{-19}$ times smaller than the electron wavelength in the previous problem.
Work Step by Step
The de Broglie wavelength is Planck’s constant/momentum, as stated on page 590.
First find the ball's momentum.
$$p = mv = (0.1 kg)(0.001 \frac{m}{s}) = 1 \times 10^{-4} \frac{kg m}{s}$$
Now find the de Broglie wavelength.
$$\lambda = \frac{h}{p} = (6.63 \times 10^{-34} J s )/(1 \times 10^{-4} \frac{kg m}{s} )$$
$$= 6.6 \times 10^{-30} m$$
This is ridiculously small, over a billion billion times smaller than the wavelength of the electron in the previous example.
