Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that the electric field for a positive charge is given by
$$\vec E=\dfrac{kq}{r^2} \;\hat r$$
where $\hat r$ is the unit vector in the direction of the electric field.
So to find the electric field at the three given points, we need to sketch them as seen below.
Hence,
$$\vec E_A=
\dfrac{kq}{r^2}\;\cos\theta_A \;\hat i+\dfrac{kq}{r^2}\;\sin\theta_A\;\hat j$$
$$\vec E_A=
\dfrac{(8.99\times 10^9)(12\times 10^{-9})}{(5\times 10^{-2})^2}\cos0^\circ\;\hat i+\\\dfrac{(8.99\times 10^9)(12\times 10^{-9})}{(5\times 10^{-2})^2}\;\sin0^\circ\;\hat j$$
$$\vec E_A=(\color{red}{\bf4.32\times 10^4}\;{\rm N/c})\hat i$$
$$\vec E_B=\dfrac{kq}{r^2}\;\cos\theta_B \;\hat i+\dfrac{kq}{r^2}\;\sin\theta_B\;\hat j$$
where $\theta_B=180^\circ-\theta$, so $\sin(180^\circ-\theta)=\sin\theta=\dfrac{y}{r}$, and hence, $\cos(180^\circ-\theta)=-\cos\theta=-\dfrac{x}{r}$.
Thus,
$$\vec E_B=\dfrac{-kq}{r^3} x\;\hat i+\dfrac{ kq}{r^3}y \;\hat j$$
recall that $r=\sqrt{x^2+y^2}$
$$\vec E_B=\dfrac{-kq}{ (x^2+y^2 )^{\frac{3}{2}}} x\;\hat i+\dfrac{ kq}{ (x^2+y^2 )^{\frac{3}{2}}} y \;\hat j$$
Plug the known;
$$\vec E_B=\dfrac{-(8.99\times 10^9)(12\times 10^{-9}) (0.05)}{ ( (-0.05)^2+ (0.05)^2 )^{\frac{3}{2}}} \;\hat i+\dfrac{(8.99\times 10^9)(12\times 10^{-9}) (0.05)}{ ( (-0.05)^2+ (0.05)^2 )^{\frac{3}{2}}} \;\hat j$$
$$\vec E_B=(\color{red}{\bf- 1.5\times 10^4}\;{\rm N/c})\hat i+(\color{red}{\bf 1.5\times 10^4}\;{\rm N/c})\hat j$$
$$\vec E_C=\dfrac{kq}{r^2}\;\cos\theta_B \;\hat i+\dfrac{kq}{r^2}\;\sin\theta_B\;\hat j$$
where $\theta_C=180^\circ+\theta$, so $\sin(180^\circ+\theta)=-\sin\theta=-\dfrac{y}{r}$, and hence, $\cos(180^\circ+\theta)=-\cos\theta=-\dfrac{x}{r}$.
Thus,
$$\vec E_C=\dfrac{-kq}{r^3} x\;\hat i+\dfrac{-kq}{r^3}y \;\hat j$$
recall that $r=\sqrt{x^2+y^2}$
$$\vec E_C=\dfrac{-kq}{ (x^2+y^2 )^{\frac{3}{2}}} x\;\hat i+\dfrac{-kq}{ (x^2+y^2 )^{\frac{3}{2}}} y \;\hat j$$
$$\vec E_C=\dfrac{-kq}{ (x^2+y^2 )^{\frac{3}{2}}} \left[ x\;\hat i+ y \;\hat j\right]$$
Plug the known;
$$\vec E_C=\dfrac{-(8.99\times 10^9)(12\times 10^{-9}) }{ ( (-0.05)^2+ (-0.05)^2 )^{\frac{3}{2}}} \left[ (0.05)\;\hat i+ (0.05) \;\hat j\right]$$
$$\vec E_C=(\color{red}{\bf- 1.5\times 10^4}\;{\rm N/c})\hat i+(\color{red}{\bf -1.5\times 10^4}\;{\rm N/c})\hat j$$
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$$\color{blue}{\bf [b]}$$
See the graph below.
