Answer
See the detailed answer below.
Work Step by Step
$\Rightarrow$ At point (1), the electric field is directly upward since the charge is positive so it is always away from the charge. And its magnitude is given by
$$E_1=\dfrac{kq}{r_1^2}$$
Plugging the known (see the given graph),
$$E_1=\dfrac{(8.99\times 10^{9})(10\times 10^{-9})}{(0.03)^2}\;\hat j=(\color{red}{\bf 1.0\times 10^5}\;{\rm N/C})\hat j$$
$\Rightarrow$ At point (2), the electric field is given by
$$E_2=\dfrac{kq}{r_2^2}\cos\theta\;\hat i+ \dfrac{kq}{r_2^2}\sin\theta\;\hat j$$
where, from the geometry of the given figure, $\cos\theta= \dfrac{0.04}{r_2}$, $\sin\theta=\dfrac{0.03}{r_2}$, and $r_2=\sqrt{0.03^2+0.04^2}$.
Thus,
$$E_2=\dfrac{kq}{r_2^2} \cdot\dfrac{0.04}{r_2}\;\hat i+ \dfrac{kq}{r_2^2} \dfrac{0.03}{r_2}\;\hat j$$
$$E_2=\dfrac{0.04kq}{r_2^3} \;\hat i+ \dfrac{0.03kq}{r_2^3} \;\hat j$$
$$E_2=\dfrac{kq}{r_2^3} \left[0.04 \;\hat i+0.03 \;\hat j\right]$$
Plugging the known (see the given graph),
$$E_2=\dfrac{(8.99\times 10^{9})(10\times 10^{-9})}{(\sqrt{0.03^2+0.04^2})^3} \left[0.04 \;\hat i+0.03 \;\hat j\right]$$
$$E_2 =(\color{red}{\bf 2.9\times 10^4}\;{\rm N/C})\hat i+(\color{red}{\bf 2.2\times 10^4}\;{\rm N/C})\hat j$$
$\Rightarrow$ At point (3), the electric field is directly eastward since the charge is positive so it is always away from the charge. And its magnitude is given by
$$E_3=\dfrac{kq}{r_3^2}$$
Plugging the known (see the given graph),
$$E_3=\dfrac{(8.99\times 10^{9})(10\times 10^{-9})}{(0.04)^2}\;\hat i=(\color{red}{\bf 5.6\times 10^4}\;{\rm N/C})\hat i$$
