Answer
See answers.
Work Step by Step
a. Treat the Earth as a uniform sphere. The angular frequency is $2 \pi$ radians per day.
$$L=I\omega=\frac{2}{5}MR_{Earth}^2(\frac{2\pi rad}{86400s})=7.1\times10^{33} kg \cdot m^2/s$$
b. Treat the Earth as a point mass. The angular frequency is one revolution per year.
$$L=I\omega=MR_{sun-Earth}^2(\frac{2\pi rad}{3.15\times10^7s})=2.7\times10^{40} kg \cdot m^2/s$$