Answer
$-6.8\times10^{5}C$
Work Step by Step
Because of the problem’s spherical symmetry, outside the Earth the electric field strength can be calculated using equation 16–4a. Solve for the charge.
$$E=k\frac{Q}{r^2}$$
$$Q=\frac{Er^2}{k}$$
$$Q=\frac{(150N/C)(6.38\times10^6m)^2}{8.988\times10^{9}(N \cdot m^2)/C^2}$$
Since the field points toward the ball, the charge must be negative.
$$Q=-6.8\times10^{5}C $$