Answer
The ratio of the Earth's rotational inertia about its own axis to the Earth's rotational inertia about the sun is $7.25\times 10^{-10}$
Work Step by Step
We can find the rotational inertia $I_a$ of the Earth about its own axis:
$I_a = \frac{2}{5}MR_E^2$
$I_a = \frac{2}{5}(5.97\times 10^{24}~kg)(6.38\times 10^6~m)^2$
$I_a = 9.72\times 10^{37}~kg~m^2$
We can find the rotational inertia $I_s$ of the Earth about the sun:
$I_s = MR^2$
$I_s = (5.97\times 10^{24}~kg)(1.50\times 10^{11}~m)^2$
$I_s = 1.34\times 10^{47}~kg~m^2$
We can find the ratio:
$\frac{I_a}{I_s} = \frac{9.72\times 10^{37}~kg~m^2}{1.34\times 10^{47}~kg~m^2} = 7.25\times 10^{-10}$
The ratio of the Earth's rotational inertia about its own axis to the Earth's rotational inertia about the sun is $7.25\times 10^{-10}$
