Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
First of all, we need to sketch the problem, as seen below.
From the geometry of the figure below, we can see that:
$\bullet$ $\theta_1=\theta_2=\theta$
$\bullet$ $r_1=r_2=\sqrt{x^2+(s/2)^2}$
$\bullet$ $\sin\theta=\dfrac{s/2}{r}=\dfrac{s}{2\sqrt{x^2+(s/2)^2}}$
$\bullet$ $\cos\theta=\dfrac{x}{r}=\dfrac{x}{\sqrt{x^2+(s/2)^2}}$
Thus, the electric field exerted by $q_1$ at a point at a distance of $x$, as seen below, is given by
$$E_1=\dfrac{kq_1}{r_1^2}\left(\cos\theta\;\hat i-\sin\theta\;\hat j\right)$$
Plug the known;
$$E_1=\dfrac{kq}{(x^2+(s/2)^2)}\left(\dfrac{x}{\sqrt{x^2+(s/2)^2}}\;\hat i-\dfrac{s}{2\sqrt{x^2+(s/2)^2}}\;\hat j\right)$$
$$E_1=\dfrac{kq}{\left(x^2+\dfrac{s^2}{4}\right)^{3/2}}\left(x\;\hat i-\dfrac{s}{2 }\;\hat j\right)\tag 1$$
And the electric field by $q_2$ at the same point is given by
$$E_2=\dfrac{kq_2}{r_2^2}\left(\cos\theta\;\hat i+\sin\theta\;\hat j\right)$$
Plug the known;
$$E_2=\dfrac{kq}{(x^2+(s/2)^2)}\left(\dfrac{x}{\sqrt{x^2+(s/2)^2}}\;\hat i+\dfrac{s}{2\sqrt{x^2+(s/2)^2}}\;\hat j\right)$$
$$E_2=\dfrac{kq}{\left(x^2+\dfrac{s^2}{4}\right)^{3/2}}\left(x\;\hat i+\dfrac{s}{2 }\;\hat j\right)\tag 2$$
Thus, the net electric field is given by
$$E_{net}=E_1+E_2$$
Plug from (1) and (2),
$$E_{net}=\dfrac{kq}{\left(x^2+\dfrac{s^2}{4}\right)^{3/2}}\left(x\;\hat i-\dfrac{s}{2 }\;\hat j\right)+\dfrac{kq}{\left(x^2+\dfrac{s^2}{4}\right)^{3/2}}\left(x\;\hat i+\dfrac{s}{2 }\;\hat j\right)$$
$$E_{net}=\dfrac{kq}{\left(x^2+\dfrac{s^2}{4}\right)^{3/2}}\left[ x\;\hat i-\dfrac{s}{2 }\;\hat j + x\;\hat i+\dfrac{s}{2 }\;\hat j\right]$$
$$\boxed{E_{net}=\left(\dfrac{2kqx}{\left(x^2+\dfrac{s^2}{4}\right)^{3/2}}\right)\;\hat i}$$
$$\color{blue}{\bf [b]}$$
Plug the known in the boxed formula above,
$$ E_{net}=\left(\dfrac{2 (9\times 10^9)(1\times 10^{-9})x}{\left(x^2+\dfrac{0.006^2}{4}\right)^{3/2}}\right)\;\hat i $$
$\Rightarrow$ At $x=0$ m,
$$ E_{}=\left(\dfrac{2 (9\times 10^9)(1\times 10^{-9})(0)}{\left(0^2+\dfrac{0.006^2}{4}\right)^{3/2}}\right)\;\hat i =\color{red}{\bf 0}\;\rm N/C$$
$\Rightarrow$ At $x=2$ mm,
$$ E_{}=\left(\dfrac{2 (9\times 10^9)(1\times 10^{-9})(0.002)}{\left(0.002^2+\dfrac{0.006^2}{4}\right)^{3/2}}\right)\;\hat i =\color{red}{\bf 768,046}\;\rm N/C$$
$\Rightarrow$ At $x=4$ mm,
$$ E_{}=\left(\dfrac{2 (9\times 10^9)(1\times 10^{-9})(0.004)}{\left(0.004^2+\dfrac{0.006^2}{4}\right)^{3/2}}\right)\;\hat i =\color{red}{\bf 567,000}\;\rm N/C$$
$\Rightarrow$ At $x=6$ mm,
$$ E_{}=\left(\dfrac{2 (9\times 10^9)(1\times 10^{-9})(0.006)}{\left(0.006^2+\dfrac{0.006^2}{4}\right)^{3/2}}\right)\;\hat i =\color{red}{\bf 357,771}\;\rm N/C$$
$\Rightarrow$ At $x=10$ mm,
$$ E_{}=\left(\dfrac{2 (9\times 10^9)(1\times 10^{-9})(0.01)}{\left(0.01^2+\dfrac{0.006^2}{4}\right)^{3/2}}\right)\;\hat i =\color{red}{\bf 158,173}\;\rm N/C$$
