How can I calculate the angular momentum of the Earth?

Hint: As a first step, you could recall the moment of inertia of a solid sphere and then substitute earth’s mass as well as radius in it to get earth’s moment of inertia. Then you could recall the angular velocity of our planet and then multiply it with the above found moment of inertia. Thus their product will give you the required answer.

Formula used:
Moment of inertia of solid sphere,
$I=\dfrac{2}{5}M{{R}^{2}}$
Angular momentum,
$L=I\omega $

Complete Step by step solution:
In the question, we are asked as to how we could find the angular momentum of our planet Earth. Since, we are given no values let us recall the known values of Earth. We know that mass and radius of earth is respectively given by,
$M=6\times {{10}^{24}}kg$
$R=6.4\times {{10}^{6}}m$

Now, let us recall the moment of inertia of a solid sphere which is given by,
$I=\dfrac{2}{5}M{{R}^{2}}$
$\Rightarrow I=\dfrac{2}{5}\left( 6\times {{10}^{24}}kg \right){{\left( 6.4\times {{10}^{6}}m \right)}^{2}}$
$\therefore I=9.72\times {{10}^{37}}kg{{m}^{2}}$ ………………………………………….. (1)

Now, recall that we have $8.64\times {{10}^{4}}s$ in a day. If $\omega $ is the angular velocity of the earth, then,
$\omega =1rev/day=\dfrac{2\pi rad}{8.64\times {{10}^{4}}s}$ …………………………………(2)

Finally, let us recall the expression for angular momentum,
$L=I\omega $

Substituting equations (1) and (2) in the above relation, we will get,
$L=9.72\times {{10}^{37}}\times \dfrac{2\pi }{8.64\times {{10}^{4}}}$
$\therefore L=7.07\times {{10}^{33}}kg{{m}^{2}}{{s}^{-1}}$

Therefore, we found the value of angular momentum of earth to be $7.07\times {{10}^{33}}kg{{m}^{2}}{{s}^{-1}}$.

Note:
Though we know that Earth actually is a spheroid we have assumed it to be a solid sphere in order to solve the question. Also, we have calculated the moment of inertia about its rotation axis. We know that earth rotates about its axis 1time every day and hence we have accordingly substituted the value of angular velocity.