#### Answer

$KE_{rot} = 2.57\times 10^{29}~J$

#### Work Step by Step

We can use $M_e = 5.97\times 10^{24}~kg$ as the mass of the earth. We can use $r = 6.38\times 10^6~m$ as the radius of the earth.
We can find the rotational inertia of the earth as:
$I = \frac{2}{5}M_e~r^2$
$I = \frac{2}{5}(5.97\times 10^{24}~kg)(6.38\times 10^6~m)^2$
$I = 9.72\times 10^{37}~kg~m^2$
We can find the angular velocity of the earth as it rotates as:
$\omega = \frac{2\pi~rad}{(24~hr)(3600~s/hr)}$
$\omega = 7.27\times 10^{-5}~rad/s$
We can find the rotational kinetic energy of the earth;
$KE_{rot} = \frac{1}{2}I\omega^2$
$KE_{rot} = \frac{1}{2}(9.72\times 10^{37}~kg~m^2)(7.27\times 10^{-5}~rad/s)^2$
$KE_{rot} = 2.57\times 10^{29}~J$