Answer
See the detailed answer below.
Work Step by Step
From the geometry of the given figure, we can see that
$r_2=1\;\rm cm$, and $r_1=r_3=\sqrt5\;\rm cm$.
The electric field is then given by
$$E=\dfrac{kq}{r^2}(\cos\theta\;\hat i+\sin\theta\;\hat j)$$
The charge here is positive so the electric field direction is away from the charge.
We chose the origin to be the charge.
$\Rightarrow$ $\textbf{At point 1:}$
$$E_1=\dfrac{kq}{r_1^2}(\cos\theta_1\;\hat i+\sin\theta_1\;\hat j)$$
where, from the geometry of the given graph, $\sin\theta_1=0.02/r_1$, and $\cos\theta_1=0.01/r_1$.
Thus,
$$E_1=\dfrac{kq}{r_1^3}(0.01\;\hat i+0.02\;\hat j)$$
Plug the given;
$$E_1=\dfrac{(8.99\times 10^9)(5\times 10^{-9})}{(\sqrt{5}\times 10^{-2})^3}(0.01\;\hat i+0.02\;\hat j)$$
$$E_1=(\color{red}{\bf 4.0\times 10^4}\;\hat i+\color{red}{\bf 8.0\times 10^4}\;\hat j)\;\rm N/C$$
$\Rightarrow$ $\textbf{At point 2:}$
So, the direction of the electric field at point 2 is $\theta_2=0^\circ$, and hence by the same approach,
$$E_2=\dfrac{kq}{r_2^2}(\cos\theta_2\;\hat i+\sin\theta_2\;\hat j)$$
$$E_2=\dfrac{kq}{r_2^2}(\cos0^\circ \;\hat i+\sin0^\circ\;\hat j)$$
$$E_2=\dfrac{kq}{r_2^2}(1 \;\hat i+0)=\dfrac{kq}{r_2^2}\;\hat i$$
Plug the given;
$$E_2=\dfrac{(8.99\times 10^9)(5\times 10^{-9})}{(1.0\times 10^{-2})^2} \;\hat i$$
$$E_2=(\color{red}{\bf 4.5\times 10^5}\;\hat i )\;\rm N/C$$
$\Rightarrow$ $\textbf{At point 3:}$
$$E_3=\dfrac{kq}{r_3^2}(\cos\theta_3\;\hat i+\sin\theta_3\;\hat j)$$
where, from the geometry of the given graph, $\sin\theta_3=-0.02/r_3$, and $\cos\theta_3=0.01/r_3$.
Thus,
$$E_3=\dfrac{kq}{r_3^3}(0.01\;\hat i-0.02\;\hat j)$$
Plug the given;
$$E_3=\dfrac{(8.99\times 10^9)(5\times 10^{-9})}{(\sqrt{5}\times 10^{-2})^3}(0.01\;\hat i-0.02\;\hat j)$$
$$E_3=(\color{red}{\bf 4.0\times 10^4}\;\hat i-\color{red}{\bf 8.0\times 10^4}\;\hat j)\;\rm N/C$$
