Answer
$-4.51\times 10^5\;\rm C$
Work Step by Step
Let's imagine a Gaussian surface that surrounds the whole earth and above its surface by a small distance.
We know, for closed surfaces, that
$$\Phi_e=\oint \vec Ed\vec A=EA=\dfrac{Q_{in}}{\epsilon_0}$$
since $E$ is uniform and constant. Recall that we are dealing with an average of the Earth's electric field that points toward its center. This means that the net charge of the earth is negative.
Thus,
$$Q_{in}=-EA_{\rm surface}\epsilon_0=-4\pi \epsilon_0R_E^2E$$
Plug the known,
$$Q_{in}= -4\pi (8.85\times 10^{-12})(6.37\times 10^6)^2(100)$$
$$Q_{in}=\color{red}{\bf -4.51\times 10^5}\;\rm C$$
