Answer
See the detailed answer below.
Work Step by Step
The charge is positive which means that the direction of the electric field is away from 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 left 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
$x=x_{charge}-x_1=1-2=\bf -1 \;\rm cm$, and
$y=y_{charge}-y_1=2+0=\bf 2\;\rm cm$
Therefore, the point position is
$${\rm Point}_1=(\color{red}{\bf -1}\;\rm cm,\color{red}{\bf 2}\;\rm cm)$$
$$\color{blue}{\bf [b]}$$
At point 2, the direction of the electric field has a positive $x$-component and a positive $y$-component too, so this point is above the charge and to the right of it.
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
$x=x_{charge}+\Delta x=1+2=\bf 3\;\rm cm$, and
$y=y_{charge}+\Delta y=2+1=\bf 3\;\rm cm$
Therefore, the point position is
$${\rm Point}_2=(\color{red}{\bf 3}\;\rm cm,\color{red}{\bf 3}\;\rm cm)$$
$$\color{blue}{\bf [c]}$$
At point 3, the direction of the electric field has a positive $x$-component and a negative $y$-component, so this point is below the charge and to the right of it.
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 clockwise with respect to $-x$-direction.
$$\theta_3=\tan^{-1}\left[\dfrac{|E_y|}{|E_x|} \right]=\tan^{-1}\left[\dfrac{|28800|}{|21600|} \right]=\bf 53.1^\circ$$
Thus,
$$\Delta x=r_3\cos\theta=5\cos 53.1^\circ=\bf 3\;\rm cm$$
$$\Delta y=r_3\sin\theta=5\sin 53.1^\circ=\bf 4\;\rm cm$$
Thus the point is at
$x=x_{charge}+\Delta x=1+3=\bf 4\;\rm cm$, and
$y=y_{charge}-\Delta y=2-4=\bf -2\;\rm cm$
Therefore, the point position is
$${\rm Point}_3=(\color{red}{\bf 4}\;\rm cm,\color{red}{\bf -2}\;\rm cm)$$
