1. Potential due to a charged arc at the center

For a small charge element $dq$ on the half ring, the potential at the center is

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

Since every element of the arc is at the same distance $R$ from the center,

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

2. Total charge on the half ring

Linear charge density is

$\lambda = 4\,\text{nC m}^{-1} = 4\times 10^{-9}\,\text{C m}^{-1}$

Radius:

$R = 10\,\text{cm} = 0.1\,\text{m}$

Length of a half ring:

$L = \pi R$

So total charge is

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

3. Substitute into potential formula

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

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

Using

$\frac{1}{4\pi\varepsilon_0} = 9\times 10^9$

we get

$V = 9\times 10^9 \times 4\times 10^{-9} \times \pi$

$V = 36\pi\,\text{V}$

4. Compare with given form

Given:

$V = x\pi\,\text{V}$

Hence,

$x = 36$

5. Comparison with stored answer

Stored correct answer = $36$

This matches the derived answer.