Answer
$1.08\times 10^5\;{\rm N/C}$
Work Step by Step
We know that the magnitude of the electric field exerted by a line of charge at distance $r$ from its center is given by
$$E=\dfrac{1}{4\pi \epsilon_0}\dfrac{|Q|}{r\sqrt{r^2+(L/2)^2} } $$
Recall example 26.3 in your textbook.
From the geometry of the figure below, we can see that the three rods are at the same distance from the center of the triangle, which means that the magnitudes of the electric fields of the three rods at this point are equal. Additionally, we can observe that the $x$-components of both $E_1$ and $E_2$ cancel each other out.
So the net electric field is given by
$$E_{net}=E_1+E_2+E_3$$
$$E_{net}=-\dfrac{1}{4\pi \epsilon_0}\dfrac{|Q|}{r\sqrt{r^2+(L/2)^2} } \cos\theta\;\hat j-\dfrac{1}{4\pi \epsilon_0}\dfrac{|Q|}{r\sqrt{r^2+(L/2)^2} } \cos\theta\;\hat j-\dfrac{1}{4\pi \epsilon_0}\dfrac{|Q|}{r\sqrt{r^2+(L/2)^2} } \;\hat j$$
$$E_{net}=-\dfrac{1}{4\pi \epsilon_0}\dfrac{|Q|}{r\sqrt{r^2+(L/2)^2} } \left[\cos\theta+\cos\theta+1\right]\;\hat j $$
$$E_{net}=-\dfrac{1}{4\pi \epsilon_0}\dfrac{|Q|}{r\sqrt{r^2+(L/2)^2} } \left[2\cos\theta+1\right]\hat j $$
where, from the geometry of the figure, $ \theta= 60^\circ$ Hence, $\cos60^\circ=\dfrac{1}{2}$,
$$E_{net}=-\dfrac{1}{4\pi \epsilon_0}\dfrac{|Q|}{r\sqrt{r^2+(L/2)^2} } \left[1+1\right]\hat j $$
$$E_{net}=-\dfrac{1}{4\pi \epsilon_0}\dfrac{2|Q|}{r\sqrt{r^2+(L/2)^2} } \hat j $$
where $\tan60^\circ=\dfrac{L/2}{r}$, so $r=L/2\tan60^\circ=\dfrac{L}{2\sqrt3}=\dfrac{10}{2\sqrt3}=5/\sqrt{3} $ cm.
Plug the known;
$$E_{net}=-\dfrac{(9\times 10^9)}{\left[\dfrac{5\sqrt{3}}{3}\times 10^{-2}\right]}\dfrac{2(10\times 10^{-9})}{ \sqrt{\left[\dfrac{5\sqrt{3}}{3}\times 10^{-2}\right]^2+\left[\dfrac{0.1}{2}\right]^2} } \hat j $$
$$E_{net}=(\color{red}{\bf -1.08\times 10^5}\;{\rm N/C})\hat j$$
