Answer
The $x$ component of $\vec{r}$ would be
$$
\frac{G m_{A} m_{D} x}{r^{3}}=-\left(\frac{4}{27}-\frac{3 \sqrt{14}}{196}\right)^{2} \frac{G m_{A}^{2}}{d^{2}}=-0.0909 \frac{G m_{A}^{2}}{d^{2}}
$$
which yields $$x=-0.0909 \frac{m_{A} r^{3}}{m_{D} d^{2}}=-0.0909 \frac{m_{A}(4.357 d)^{3}}{\left(4 m_{A}\right) d^{2}}=-1.88 d$$
Work Step by Step
The $x$ component of $\vec{r}$ would be
$$
\frac{G m_{A} m_{D} x}{r^{3}}=-\left(\frac{4}{27}-\frac{3 \sqrt{14}}{196}\right)^{2} \frac{G m_{A}^{2}}{d^{2}}=-0.0909 \frac{G m_{A}^{2}}{d^{2}}
$$
which yields $$x=-0.0909 \frac{m_{A} r^{3}}{m_{D} d^{2}}=-0.0909 \frac{m_{A}(4.357 d)^{3}}{\left(4 m_{A}\right) d^{2}}=-1.88 d$$
