1. Potential due to a charged arc at its centre

For a small element $dq$ on a ring/arc of radius $R$, the potential at the centre is

$dV = \frac{1}{4\pi \varepsilon_0} \frac{dq}{R}$

Since every point of the arc is at the same distance $R$ from the centre,

$V = \frac{1}{4\pi \varepsilon_0} \frac{1}{R} \int dq = \frac{1}{4\pi \varepsilon_0} \frac{Q}{R}$

2. Charge on each half ring

A half ring of radius $R$ has length

$L = \pi R$

Given linear charge density $\lambda$,

$Q = \lambda L = \lambda \pi R$

Hence potential at the centre due to one half ring is

$V = \frac{1}{4\pi \varepsilon_0} \frac{\lambda \pi R}{R} = \frac{\lambda}{4\varepsilon_0}$

So, importantly, the potential due to a half ring is independent of radius.

3. Potential due to two concentric half rings

Each half ring contributes

$\frac{\lambda}{4\varepsilon_0}$

Therefore total potential at the common centre is

$V_{\text{total}} = \frac{\lambda}{4\varepsilon_0} + \frac{\lambda}{4\varepsilon_0} = \frac{\lambda}{2\varepsilon_0}$

4. Matching with options

$\boxed{V = \frac{\lambda}{2\varepsilon_0}}$

So the correct option is:

A

5. Comparison with stored answer

Stored correct answer: A

My derived answer: A

They agree.