1. Given data

   • Main scale reading (MSR) $= 1\,\text{mm}$
   • Circular scale reading (CSR) $= 42$ divisions
   • Pitch $= 1\,\text{mm}$
   • Number of circular scale divisions $= 100$

2. Find least count of screw gauge $\text{Least Count} = \frac{\text{Pitch}}{\text{Number of circular divisions}} = \frac{1}{100}\,\text{mm} = 0.01\,\text{mm}$

3. Find circular scale contribution $\text{CSR contribution} = 42 \times 0.01 = 0.42\,\text{mm}$

4. Total diameter of wire $\text{Diameter} = \text{MSR} + \text{CSR contribution} = 1 + 0.42 = 1.42\,\text{mm}$

5. Compare with given form Diameter is given as $\frac{x}{50}\,\text{mm}$ So, $\frac{x}{50} = 1.42$ $x = 1.42 \times 50 = 71$

6. Check options

   • A: $42$ ❌
   • B: $71$ ✅
   • C: $21$ ❌
   • D: $142$ ❌

Therefore, the correct answer is B: 71.