Answer
See the detailed answer below.
Work Step by Step
First, we need to draw these 3 dots and the electric field direction. We know that the electric field direction is away from a positive charge and is toward the negative charge, see the second figure below.
We know that the electric field exerted on a point at distance $r$ from the center of a long charged rod is given by
$$E=\dfrac{1}{4\pi \epsilon_0}\dfrac{|Q|}{r\sqrt{r^2+\left(\frac{L}{2}\right)^2}}$$
We have two glass rods, we name one of them, the upper, rod A, and the other rod B.
Thus, at point 1, the net electric field is given by
$$E_1=E_{A}+E_{B}$$
We chose upward to be our positive direction,
$$E_1=-\dfrac{1}{4\pi \epsilon_0} \dfrac{|Q_{A}|}{r_{A}\sqrt{r_{A}^2+
\left(\frac{L_{A}}{2}\right)^2}}\;\hat j+\dfrac{1}{4\pi \epsilon_0}\dfrac{|Q_{B}|}{r_{B}\sqrt{r_{B}^2+\left(\frac{L_{B}}{2}\right)^2}}\;\hat j $$
Recall that $|Q_{A}|=|Q_{B}|=Q$, and $L_{A}=L_{B}=L$
$$E_1=-\dfrac{Q}{4\pi \epsilon_0} \left[ \dfrac{1}{r_{A}\sqrt{r_{A}^2+\left(\frac{L}{2}\right)^2}}\;-\dfrac{1}{r_{B}\sqrt{r_{B}^2+\left(\frac{L}{2}\right)^2}}\right]\hat j\tag 1$$
Noting that $r_{A}=r_{A\rightarrow P_1}=0.01$ m while $r_{B}=r_{B\rightarrow P_1}=0.03$ m,
$$E_1=-\dfrac{(10\times 10^{-9})}{4\pi (8.85\times 10^{-12})} \left[ \dfrac{1}{(0.01)\sqrt{(0.01)^2+\left(\frac{0.1}{2}\right)^2}}-\dfrac{1}{(0.03)\sqrt{(0.03)^2+\left(\frac{0.1}{2}\right)^2}}\right]\hat j $$
$$E_1=(\color{red}{\bf -1.25\times 10^5}\;\hat j)\;\rm N/C$$
By the same approach, using (1), at point 2, the net electric field is given by
$$E_2=-\dfrac{Q}{4\pi \epsilon_0} \left[ \dfrac{1}{r_{A}\sqrt{r_{A}^2+\left(\frac{L}{2}\right)^2}}-\dfrac{1}{r_{B}\sqrt{r_{B}^2+\left(\frac{L}{2}\right)^2}}\right]\hat j $$
Noting that $r_{A}=r_{A\rightarrow P_2}=0.02$ m while $r_{B}=r_{B\rightarrow P_2}=0.02$ m,
$$E_2=-\dfrac{(10\times 10^{-9})}{4\pi (8.85\times 10^{-12})} \left[ \dfrac{1}{(0.02)\sqrt{(0.02)^2+\left(\frac{0.1}{2}\right)^2}}-\dfrac{1}{(0.02)\sqrt{(0.02)^2+\left(\frac{0.1}{2}\right)^2}}\right]\hat j $$
$$E_2=(\color{red}{\bf 0}\;\hat j)\;\rm N/C$$
By the same approach, using (1), at point 3, the net electric field is given by
$$E_3=-\dfrac{Q}{4\pi \epsilon_0} \left[ \dfrac{1}{r_{A}\sqrt{r_{A}^2+\left(\frac{L}{2}\right)^2}}\;+\dfrac{1}{r_{B}\sqrt{r_{B}^2+\left(\frac{L}{2}\right)^2}}\right]\hat j $$
Noting that $r_{A}=r_{A\rightarrow P_3}=0.03$ m while $r_{B}=r_{B\rightarrow P_3}=0.01$ m,
$$E_3=-\dfrac{(10\times 10^{-9})}{4\pi (8.85\times 10^{-12})} \left[ \dfrac{1}{(0.03)\sqrt{(0.03)^2+\left(\frac{0.1}{2}\right)^2}}-\dfrac{1}{(0.01)\sqrt{(0.01)^2+\left(\frac{0.1}{2}\right)^2}}\right]\hat j $$
$$E_3=(\color{red}{\bf 1.25\times 10^5}\;\hat j)\;\rm N/C$$