First, calculate the radius of the ring. The wire of length \( l = 0.1\pi \) m is bent into a ring, so the circumference of the ring is:
\[
2\pi r = 0.1\pi \quad \Rightarrow \quad r = \frac{0.1\pi}{2\pi} = 0.05 \, \text{m}
\]
Next, calculate the mass of the ring. The linear density is \( \lambda = \frac{1}{\pi} \) kg m\(^{-1}\), so:
\[
m = \lambda \times l = \frac{1}{\pi} \times 0.1\pi = 0.1 \, \text{kg}
\]
The moment of inertia of a ring about an axis passing through its center and perpendicular to its plane is:
\[
I = m r^2
\]
Substituting the values:
\[
I = 0.1 \times (0.05)^2 = 0.1 \times 0.0025 = 0.00025 \, \text{kg m}^2 = 0.25 \times 10^{-3} \, \text{kg m}^2
\]
So, the moment of inertia of the ring is 0.25 \(\times 10^{-3}\) kg m\(^2\).