Answer
$-5.11\times10^{-11}C$
Work Step by Step
Because of the problem’s spherical symmetry, outside the ball the electric field strength can be calculated using equation 16–4a. Solve for the charge.
$$E=k\frac{Q}{r^2}$$
$$Q=\frac{Er^2}{k}$$
$$Q=\frac{(3.75\times10^2N/C)(0.0350m)^2}{8.988\times10^{9}(N \cdot m^2)/C^2}$$
Since the field points toward the ball, the charge must be negative.
$$Q=-5.11\times10^{-11}C$$