1. The resistance of a wire is

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

where $\rho$ is resistivity, $L$ is length, and $A$ is cross-sectional area.

2. Since the wire is stretched and its volume remains constant, we have

$AL = \text{constant}$

So if length increases, area decreases accordingly.

3. Let the original length be $L$ and area be $A$.

Given the wire is made $0.5\%$ longer,

$\frac{\Delta L}{L} = 0.5\% = 0.005$

Thus,

$L' = L(1+0.005) = 1.005L$

4. From constant volume:

$A'L' = AL$

So,

$A' = \frac{AL}{L'} = \frac{A}{1.005}$

5. New resistance:

$R' = \rho \frac{L'}{A'} = \rho \frac{1.005L}{A/1.005} = \rho \frac{L}{A}(1.005)^2$

Hence,

$\frac{R'}{R} = (1.005)^2$

6. Percentage change in resistance:

$\frac{\Delta R}{R} = (1.005)^2 - 1$

$= 1.010025 - 1 = 0.010025$

So,

$\%\text{ change in }R = 0.010025 \times 100 = 1.0025\%$

This is approximately

$1.0\%$

7. Therefore, the correct option is:

C: $1.0\%$

8. Comparison with stored answer:

Stored correct answer is C, which matches the derived answer.