Step 1: Understand the question.
A half ring (semicircle) of radius $r$ carries charge spread evenly along it, with linear charge density $\lambda$. We must find the electric field at the centre of the circle.
Step 2: Break the ring into tiny pieces.
Take a very small piece of the arc that makes a tiny angle $d\theta$ at the centre. Its length is $r\,d\theta$, so its charge is:
\[ dq = \lambda\, r\, d\theta \]
Step 3: Field from one tiny piece.
Each tiny charge is at distance $r$ from the centre, so it makes a small field:
\[ dE = \frac{1}{4\pi\epsilon_0}\,\frac{dq}{r^2} = \frac{1}{4\pi\epsilon_0}\,\frac{\lambda\, d\theta}{r} \]
Step 4: Use symmetry.
The half ring is balanced about its middle line. For every piece on one side there is a matching piece on the other side. The sideways parts of their fields cancel out. Only the parts pointing along the symmetry line (call it the $y$ direction) add up. We measure each piece by an angle $\theta$ from this line, so we keep $dE\cos\theta$.
Step 5: Add up (integrate) all the pieces.
The angle runs from $-\tfrac{\pi}{2}$ to $+\tfrac{\pi}{2}$ for a semicircle:
\[ E = \int_{-\pi/2}^{\pi/2} \frac{\lambda}{4\pi\epsilon_0 r}\cos\theta\, d\theta = \frac{\lambda}{4\pi\epsilon_0 r}\big[\sin\theta\big]_{-\pi/2}^{\pi/2} \]
Step 6: Put in the limits.
Since $\sin(\tfrac{\pi}{2}) = 1$ and $\sin(-\tfrac{\pi}{2}) = -1$:
\[ E = \frac{\lambda}{4\pi\epsilon_0 r}\,[1 - (-1)] = \frac{\lambda}{4\pi\epsilon_0 r}\times 2 \]
\[ E = \frac{\lambda}{2\pi\epsilon_0 r} \]
This matches option (1).
\[ \boxed{E = \frac{\lambda}{2\pi\epsilon_0 r}} \]