Answer
$q=1.0 \times 10^{-11} \mathrm{C} .$
Work Step by Step
To calculate the magnitude of the charge, we note that the field magnitude measured at $(2.0 \mathrm{~cm}, 0)$ (which is $r=0.030 \mathrm{~m}$ from the charge) is
$
|\vec{E}|=\frac{1}{4 \pi \varepsilon_0} \frac{q}{r^2}=100 \mathrm{~N} / \mathrm{C} .
$
Therefore,
$
q=4 \pi \varepsilon_0|\vec{E}| r^2=\frac{(100 \mathrm{~N} / \mathrm{C})(0.030 \mathrm{~m})^2}{8.99 \times 10^9 \mathrm{~N} \cdot \mathrm{m}^2 / \mathrm{C}^2}=1.0 \times 10^{-11} \mathrm{C} .
$
