Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
First, we need to sketch the problem as shown below.
The electric field exerted by the small black segment of the rod, which has a charge of $dq$ and a length of $ds$, at the center of the circle (which we chose as our origin), is $dE$.
$$dE=\dfrac{1}{4\pi \epsilon_0}\dfrac{dq}{R^2}(\cos\theta\;\hat i+\sin\theta\;\hat j) \tag 1$$
Let's assume that the linear charge density of the rod is $\lambda$, so
$$\lambda=\dfrac{Q}{L}=\dfrac{dq}{ds}$$
where $ds$ is the length of the small arc (black segment in the figure below) and we know that the arc length is given by $R\theta$ where $\theta$ in radians, so
$$\lambda=\dfrac{Q}{L}=\dfrac{dq}{Rd\theta}$$
Hence,
$$dq=\dfrac{QRd\theta}{L}$$
Plug into (1),
$$dE=\dfrac{1}{4\pi \epsilon_0}\dfrac{QRd\theta}{LR^2}(\cos\theta\;\hat i+\sin\theta\;\hat j) $$
$$dE=\dfrac{Q}{(4\pi \epsilon_0)LR }(\cos\theta d\theta\;\hat i+\sin\theta d\theta\;\hat j) $$
Now we need to find the radius $R$ in terms of $L$, where we know that the length of a quarter-circle is given by $L=\pi R/2$, so $R=2L/\pi$
$$\boxed{dE=\dfrac{\pi Q}{(4\pi \epsilon_0)2L^2 }(\cos\theta d\theta\;\hat i+\sin\theta d\theta\;\hat j) }$$
And in terms of $R$,
$$\boxed{dE=\dfrac{2 Q}{(4\pi \epsilon_0)\pi R^2 }(\cos\theta d\theta\;\hat i+\sin\theta d\theta\;\hat j) }$$
$$\color{blue}{\bf [b]}$$
Now to find the net electric field of the whole semi-circle rod, we need to integrate both sides relative to $\theta$ where $\theta$ varies from $0$ to $\pi/2$.
$$E=\int dE=\dfrac{2 Q}{(4\pi \epsilon_0)\pi R^2 }\left[\int_0^{\pi/2} \cos\theta d\theta\;\hat i+\int_0^{\pi/2}\sin\theta d\theta\;\hat j\right] $$
$$\color{blue}{\bf [c]}$$
$$E =\dfrac{2 Q}{(4\pi \epsilon_0)\pi R^2 }\left[(\sin(\pi/2)-\sin 0)\;\hat i- (\cos(\pi/2)-\cos0)\;\hat j\right] $$
$$\boxed{\vec E =\dfrac{2 Q}{(4\pi \epsilon_0)\pi R^2 }\left[\hat i+\hat j\right] }$$