1. Given data

   • Length of wire: $L = 25\,\text{m}$
   • Cross-sectional area: $A = 5\,\text{mm}^2 = 5 \times 10^{-6}\,\text{m}^2$
   • Resistivity: $\rho = 2 \times 10^{-6}\,\Omega\text{m}$

2. Resistance of the full wire

   Using $R = \frac{\rho L}{A}$

   $R = \frac{(2 \times 10^{-6})(25)}{5 \times 10^{-6}}$

   $R = \frac{50 \times 10^{-6}}{5 \times 10^{-6}} = 10\,\Omega$

   So, the total resistance of the wire is $R_{\text{total}} = 10\,\Omega$

3. Wire bent into a complete circle

   Diametrically opposite points divide the circular wire into two equal halves.

   Therefore, each half has resistance $R_{\text{half}} = \frac{10}{2} = 5\,\Omega$

4. Equivalent resistance between diametrically opposite points

   These two halves are connected in parallel between the diametrically opposite points.

   Hence, $R_{\text{eq}} = \frac{5 \times 5}{5 + 5} = \frac{25}{10} = 2.5\,\Omega$

5. Check options

   • A: $100\,\Omega$ ❌
   • B: $2.5\,\Omega$ ✅
   • C: $12.5\,\Omega$ ❌
   • D: $50\,\Omega$ ❌

Therefore, the correct answer is Option B.