(N/A) For the electron:
Mass $m = 9.11 \times 10^{-31} \; kg$,speed $v = 5.4 \times 10^{6} \; m/s$.
Momentum $p = mv = (9.11 \times 10^{-31} \; kg) \times (5.4 \times 10^{6} \; m/s) = 4.92 \times 10^{-24} \; kg \cdot m/s$.
de Broglie wavelength $\lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34} \; J \cdot s}{4.92 \times 10^{-24} \; kg \cdot m/s} \approx 1.35 \times 10^{-10} \; m = 0.135 \; nm$.
$(b)$ For the ball:
Mass $m' = 0.150 \; kg$,speed $v' = 30.0 \; m/s$.
Momentum $p' = m'v' = (0.150 \; kg) \times (30.0 \; m/s) = 4.50 \; kg \cdot m/s$.
de Broglie wavelength $\lambda' = \frac{h}{p'} = \frac{6.63 \times 10^{-34} \; J \cdot s}{4.50 \; kg \cdot m/s} \approx 1.47 \times 10^{-34} \; m$.
The de Broglie wavelength of the electron is comparable to $X$-ray wavelengths,whereas for the ball,it is extremely small (about $10^{-19}$ times the size of a proton),making it beyond experimental measurement.