Answer
See the detailed answer below.
Work Step by Step
The charge is negative which means that the direction of the electric field is toward the charge itself.
$$\color{blue}{\bf [a]}$$
At point 1, the direction of the electric field is toward the negative $x$-direction, and it has no $y$-component, so this point is to the right of the charge.
$$E_1=\dfrac{kq}{r_1^2}(\cos\theta_1\;\hat i+\sin\theta_1\;\hat j)$$
where $\theta_1=180^\circ$, so
$$E_1=\dfrac{kq}{r_1^2}(\cos180^\circ\;\hat i+\sin180^\circ\;\hat j)=\dfrac{-kq}{r_1^2}\;\hat i$$
Note that $r^2=x_1^2+y_1^2=x_1^2+0=x_1^2$,
$$E_1 =\dfrac{-kq}{x_1^2} \;\hat i$$
Thus,
$$x_1=\sqrt{\dfrac{-kq}{E_1 }}$$
Plug the known;
$$x_1=\sqrt{\dfrac{-(9.0\times 10^9)(10\times 10^{-9})}{-2.25\times 10^5}}=\;0.02\;\rm m=\bf 2.0\;cm$$
Thus the point is at
$x=x_{charge}+x_1=2+2=\bf 4 \;\rm cm$, and
$y=y_{charge}+y_1=1+0=\bf 1\;\rm cm$
Therefore, the point position is
$${\rm Point}_1=(\color{red}{\bf 4}\;\rm cm,\color{red}{\bf 1}\;\rm cm)$$
$$\color{blue}{\bf [b]}$$
At point 2, the direction of the electric field has a positive $x$-component and a negative $y$-component, so this point is above the charge and to the left of it. Hence, it is in the first quadrant with respect to the charge.
But it is in the fourth quadrant with respect to point 2 itself.
The magnitude of this electric field is given by
$$E_2=\dfrac{kq}{r_2^2}=\sqrt{E_x^2+E_y^2} $$
Solving for $r_2$,
$$r_2^2=\dfrac{kq}{\sqrt{E_x^2+E_y^2}} $$
$$r_2 =\sqrt{\dfrac{kq}{\sqrt{E_x^2+E_y^2}} }$$
Plug the known;
$$r_2 =\sqrt{\dfrac{(9.0\times 10^9)(10\times 10^{-9})}{\sqrt{(161000)^2+(80500)^2}} }=0.0224\;\rm m=\bf 2.24\;cm$$
This is the direct distance between point 2 and the charge, and its direction is given by
$$\tan\theta=\dfrac{|E_y|}{|E_x|} $$
where $\theta$ here is clockwise with respect to $+x$-direction.
$$\theta=\tan^{-1}\left[\dfrac{|E_y|}{|E_x|} \right]=\tan^{-1}\left[\dfrac{|80500|}{|161000|} \right]=\bf 25.57^\circ$$
Thus,
$$\Delta x=r_2\cos\theta=2.24\cos 25.57^\circ=\bf 2.0\;\rm cm$$
$$\Delta y=r_2\sin\theta=2.24\sin25.57^\circ=\bf 1.0\;\rm cm$$
Thus the point is at
$x=x_{charge}-\Delta x=2-2=\bf 0\;\rm cm$, and
$y=y_{charge}+\Delta y=1+1=\bf 2\;\rm cm$
Therefore, the point position is
$${\rm Point}_2=(\color{red}{\bf 0}\;\rm cm,\color{red}{\bf 2}\;\rm cm)$$
$$\color{blue}{\bf [c]}$$
At point 3, the direction of the electric field has a positive $x$-component and a positive $y$-component too, so this point is below the charge and to the left of it. Hence this point is in the third quadrant with respect to itself.
The magnitude of this electric field is given by
$$E_3=\dfrac{kq}{r_3^2}=\sqrt{E_x^2+E_y^2} $$
Solving for $r_3$,
$$r_3^2=\dfrac{kq}{\sqrt{E_x^2+E_y^2}} $$
$$r_3 =\sqrt{\dfrac{kq}{\sqrt{E_x^2+E_y^2}} }$$
Plug the known;
$$r_3 =\sqrt{\dfrac{(9.0\times 10^9)(10\times 10^{-9})}{\sqrt{(28800)^2+(21600)^2}} }=0.05\;\rm m=\bf 5.0\;cm$$
This is the direct distance between point 3 and the charge, and its direction is given by
$$\tan\theta_3=\dfrac{|E_y|}{|E_x|} $$
where $\theta_3$ here is counterclockwise with respect to $+x$-direction.
$$\theta_3=\tan^{-1}\left[\dfrac{|E_y|}{|E_x|} \right]=\tan^{-1}\left[\dfrac{|21600|}{|28800|} \right]=\bf 36.9^\circ$$
Thus,
$$\Delta x=r_3\cos\theta=5\cos 36.9^\circ=\bf 4.0\;\rm cm$$
$$\Delta y=r_3\sin\theta=5\sin 36.9^\circ=\bf 3.0\;\rm cm$$
Thus the point is at
$x=x_{charge}-\Delta x=2-4=\bf -2\;\rm cm$, and
$y=y_{charge}-\Delta y=1-3=\bf -2\;\rm cm$
Therefore, the point position is
$${\rm Point}_3=(\color{red}{\bf -2}\;\rm cm,\color{red}{\bf -2}\;\rm cm)$$
