In a screw gauge,there are `100` divisions on the circular scale and the main scale moves by `0.5mm` on a complete rotation of the circular scale.The zero of circular scale lies `6` divisions below the line of graduation when two studs are brought in contact with each other.When a wire is placed between the stds, `4` linear scale divisions are clearly visible while `46` th of vision the circular scale coincide with the reference line.The diameter of the wire is `.....times10^(-2)mm`.

Answer

To find the diameter of the wire using the screw gauge, we will follow these steps:

### Step 1: Calculate the Least Count (LC)
The least count of the screw gauge is calculated using the formula:
\[ \text{LC} = \frac{\text{Pitch}}{\text{Number of divisions on circular scale}} \]
Given:
- Pitch = 0.5 mm
- Number of divisions on circular scale = 100

So,
\[ \text{LC} = \frac{0.5 \, \text{mm}}{100} = 0.005 \, \text{mm} \]

### Step 2: Determine the Zero Error
The zero of the circular scale lies 6 divisions below the line of graduation when the two studs are in contact. This indicates a positive zero error. The zero error can be calculated as:
\[ \text{Zero Error} = \text{Number of divisions below} \times \text{LC} \]
\[ \text{Zero Error} = 6 \times 0.005 \, \text{mm} = 0.03 \, \text{mm} \]

### Step 3: Calculate the Measured Thickness
When the wire is placed between the studs:
- The main scale reading (MSR) is given by the visible linear scale divisions. Here, 4 linear scale divisions are visible.
- Each linear scale division is 0.5 mm, so:
\[ \text{MSR} = 4 \times 0.5 \, \text{mm} = 2.0 \, \text{mm} \]
- The circular scale reading (CSR) is 46 divisions. Thus:
\[ \text{CSR} = 46 \times \text{LC} = 46 \times 0.005 \, \text{mm} = 0.23 \, \text{mm} \]
- Therefore, the measured thickness (MT) can be calculated as:
\[ \text{Measured Thickness} = \text{MSR} + \text{CSR} = 2.0 \, \text{mm} + 0.23 \, \text{mm} = 2.23 \, \text{mm} \]

### Step 4: Calculate the Actual Thickness
The actual thickness can be found by subtracting the zero error from the measured thickness:
\[ \text{Actual Thickness} = \text{Measured Thickness} - \text{Zero Error} \]
\[ \text{Actual Thickness} = 2.23 \, \text{mm} - 0.03 \, \text{mm} = 2.20 \, \text{mm} \]

### Step 5: Convert to Required Format
To express the diameter in the required format:
\[ \text{Diameter} = 2.20 \, \text{mm} = 220 \times 10^{-2} \, \text{mm} \]

### Final Answer
The diameter of the wire is:
\[ \boxed{220} \]