Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that the electric field due to a continuous charge distribution is given by
$$E=\dfrac{1}{4\pi \epsilon_0}\int_0^r\dfrac{1}{r^2}dq$$
Let's assume that the linear charge density of this rod is $\lambda$ which is given by
$$\lambda=\dfrac{Q}{L}=\dfrac{dq}{dx}$$
where $dL$ is the length of the small black part in the figure below and $dx$ is its length.
Thus,
$$dq=\dfrac{Qdx}{L}\tag 1$$
So the electric field due to this small segment, as seen in the figure below, is given by
$$dE=\dfrac{1}{4\pi \epsilon_0} \dfrac{dq}{(r-x)^2}$$
For the electric field from the rod, we need to take the integral for both sides,
$$E_x=\dfrac{1}{4\pi \epsilon_0}\int_{-L/2}^{L/2} \dfrac{dq}{(r-x)^2}$$
Plug $dx$ from (1),
$$E_x=\dfrac{Q/L}{4\pi \epsilon_0}\int_{-L/2}^{L/2} \dfrac{1}{(r-x)^2}dx$$
$$E_x=\dfrac{Q/L}{4\pi \epsilon_0}\bigg[ \dfrac{1}{(r-x) }\bigg]_{-L/2}^{L/2}$$
$$E_x=\dfrac{Q/L}{4\pi \epsilon_0}\bigg[\dfrac{1}{r-(L/2)}-\dfrac{1}{r+(L/2)}\bigg] $$
$$E_x=\dfrac{Q/L}{4\pi \epsilon_0}\bigg[\dfrac{r+(L/2)-[r-(L/2)]}{r^2-(L/2)^2} \bigg] $$
$$E_x=\dfrac{Q/L}{4\pi \epsilon_0}\bigg[\dfrac{ L}{r^2-(L/2)^2} \bigg] $$
$$E_x=\dfrac{Q }{4\pi \epsilon_0}\bigg[\dfrac{1}{r^2-(L/2)^2} \bigg] $$
$$\boxed{E_x=\dfrac{1}{4\pi \epsilon_0}\left[\dfrac{Q}{r^2-\frac{L^2}{4}} \right] }$$
$$\color{blue}{\bf [b]}$$
when $r\gt \gt L$, then $r^2-\dfrac{L^2}{4}\approx r^2$.
Thus,
$$E_x=\dfrac{1 }{4\pi \epsilon_0}\bigg[\dfrac{Q}{r^2 } \bigg] $$
which is acceptable since at a long distance from the rod, it appears as a point charge.
$$\color{blue}{\bf [c]}$$
Plug the known into the boxed formula above,
$$E_x=\dfrac{(9.0\times 10^9) (3\times 10^{-9})}{0.03^2-0.025^2}$$
$$E_x=\color{red}{\bf 9.82\times 10^4}\;\rm N/C$$