Answer
$x_0=-1.0 \mathrm{~cm}$.
Work Step by Step
The fact that the second measurement at the location $(2.0 \mathrm{~cm}, 0)$ gives $\vec{E}=(100 \mathrm{~N} / \mathrm{C}) \hat{\mathrm{i}}$ indicates that $y_0=0$, that is, the charge must be somewhere on the $x$ axis. Thus, the above expression can be simplified to
$
\vec{E}=\frac{q}{4 \pi \varepsilon_0} \frac{\left(x-x_0\right) \hat{i}+y \hat{j}}{\left[\left(x-x_0\right)^2+y^2\right]^{3 / 2}} .
$
On the other hand, the field at $(3.0 \mathrm{~cm}, 3.0 \mathrm{~cm})$ is $\vec{E}=(7.2 \mathrm{~N} / \mathrm{C})(4.0 \hat{\mathrm{i}}+3.0 \hat{\mathrm{j}})$, which gives $E_y / E_x=3 / 4$. Thus, we have
$
\frac{3}{4}=\frac{3.0 \mathrm{~cm}}{3.0 \mathrm{~cm}-x_0}
$
which implies $x_0=-1.0 \mathrm{~cm}$.
