A rod of length L carries a total charge Q distributed uniformly...

Let the radius of the semicircle be $R$. Since the rod of length $L$ is bent into a semicircle, the length of the arc is equal to $L$.

$\pi R = L \Rightarrow R = \dfrac{L}{\pi}$

The linear charge density of the rod is $\lambda = \dfrac{Q}{L}$.

Consider a small element of the semicircle subtending an angle $d\theta$ at the centre. The charge on this element is $dq = \lambda R d\theta$.

The electric field due to this element at the centre is $dE = \dfrac{1}{4\pi\varepsilon_0} \dfrac{dq}{R^2} = \dfrac{1}{4\pi\varepsilon_0} \dfrac{\lambda R d\theta}{R^2} = \dfrac{\lambda}{4\pi\varepsilon_0 R} d\theta$.

By symmetry, the components of the electric field perpendicular to the angle bisector cancel out. The net electric field is along the angle bisector.

$E = \int_{-\pi/2}^{\pi/2} dE \cos\theta = \int_{-\pi/2}^{\pi/2} \dfrac{\lambda}{4\pi\varepsilon_0 R} \cos\theta d\theta$

$E = \dfrac{\lambda}{4\pi\varepsilon_0 R} [\sin\theta]_{-\pi/2}^{\pi/2} = \dfrac{\lambda}{4\pi\varepsilon_0 R} (1 - (-1)) = \dfrac{2\lambda}{4\pi\varepsilon_0 R} = \dfrac{\lambda}{2\pi\varepsilon_0 R}$

Substituting $\lambda = \dfrac{Q}{L}$ and $R = \dfrac{L}{\pi}$ into the expression for $E$:

$E = \dfrac{Q/L}{2\pi\varepsilon_0 (L/\pi)} = \dfrac{Q \pi}{2\pi\varepsilon_0 L^2} = \dfrac{Q}{2\varepsilon_0 L^2}$