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 exrted on a point at distance $r$ from the center of a long charged rod is given by
$$E=\dfrac{1}{4\pi \epsilon_0}\dfrac{|Q|}{r\sqrt{r^2+\left(\frac{L}{2}\right)^2}}$$
Thus, at point 1, the net electric field is given by
$$E_1=E_{glass}+E_{plastic}$$
We chose upward to be our positive direction,
$$E_1=-\dfrac{1}{4\pi \epsilon_0} \dfrac{|Q_{glass}|}{r_{glass}\sqrt{r_{glass}^2+
\left(\frac{L_{glass}}{2}\right)^2}}\;\hat j-\dfrac{1}{4\pi \epsilon_0}\dfrac{|Q_{plastic}|}{r_{plastic}\sqrt{r_{plastic}^2+\left(\frac{L_{plastic}}{2}\right)^2}}\;\hat j $$
Recall that $|Q_{glass}|=|Q_{plastic}|=Q$, and $L_{glass}=L_{plastic}=L$
$$E_1=-\dfrac{Q}{4\pi \epsilon_0} \left[ \dfrac{1}{r_{glass}\sqrt{r_{glass}^2+\left(\frac{L}{2}\right)^2}}\;+\dfrac{1}{r_{plastic}\sqrt{r_{plastic}^2+\left(\frac{L}{2}\right)^2}}\right]\hat j\tag 1$$
Noting that $r_{glass}=r_{glass\rightarrow P_1}=0.01$ m while $r_{plastic}=r_{plastic\rightarrow P_1}=0.03$ m,
$$E_1=-\dfrac{(10\times 10^{-9})}{4\pi (8.85\times 10^{-12})} \left[ \dfrac{1}{(0.01)\sqrt{(0.01)^2+\left(\frac{0.1}{2}\right)^2}}+\dfrac{1}{(0.03)\sqrt{(0.03)^2+\left(\frac{0.1}{2}\right)^2}}\right]\hat j $$
$$E_1=(\color{red}{\bf -2.28\times 10^5}\;\hat j)\;\rm N/C$$
By the same approach, using (1), at point 2, the net electric field is given by
$$E_2=-\dfrac{Q}{4\pi \epsilon_0} \left[ \dfrac{1}{r_{glass}\sqrt{r_{glass}^2+\left(\frac{L}{2}\right)^2}}\;+\dfrac{1}{r_{plastic}\sqrt{r_{plastic}^2+\left(\frac{L}{2}\right)^2}}\right]\hat j $$
Noting that $r_{glass}=r_{glass\rightarrow P_2}=0.02$ m while $r_{plastic}=r_{plastic\rightarrow P_2}=0.02$ m,
$$E_2=-\dfrac{(10\times 10^{-9})}{4\pi (8.85\times 10^{-12})} \left[ \dfrac{1}{(0.02)\sqrt{(0.02)^2+\left(\frac{0.1}{2}\right)^2}}+\dfrac{1}{(0.02)\sqrt{(0.02)^2+\left(\frac{0.1}{2}\right)^2}}\right]\hat j $$
$$E_2=(\color{red}{\bf -1.67\times 10^5}\;\hat j)\;\rm N/C$$
By the same approach, using (1), at point 3, the net electric field is given by
$$E_3=-\dfrac{Q}{4\pi \epsilon_0} \left[ \dfrac{1}{r_{glass}\sqrt{r_{glass}^2+\left(\frac{L}{2}\right)^2}}\;+\dfrac{1}{r_{plastic}\sqrt{r_{plastic}^2+\left(\frac{L}{2}\right)^2}}\right]\hat j $$
Noting that $r_{glass}=r_{glass\rightarrow P_3}=0.03$ m while $r_{plastic}=r_{plastic\rightarrow P_3}=0.01$ m,
$$E_2=-\dfrac{(10\times 10^{-9})}{4\pi (8.85\times 10^{-12})} \left[ \dfrac{1}{(0.02)\sqrt{(0.02)^2+\left(\frac{0.1}{2}\right)^2}}+\dfrac{1}{(0.02)\sqrt{(0.02)^2+\left(\frac{0.1}{2}\right)^2}}\right]\hat j $$
$$E_3=(\color{red}{\bf -2.28\times 10^5}\;\hat j)\;\rm N/C$$
