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**Hint:**The Poisson’s ratio for a wire is given. When the wire is stretched, the length increases by 0.05%. Now by definition, we know the Poisson’s ratio $\sigma = \dfrac{{{\text{lateral strain}}}}{{{\text{longitudinal strain}}}}$. Longitudinal strain is given, so we can find the lateral strain by substituting the given values.

**Formula used:**

Poisson’s ratio $\sigma = \dfrac{{{\text{lateral strain}}}}{{{\text{longitudinal strain}}}}$ $ \Rightarrow \sigma = \dfrac{{\dfrac{{\Delta D}}{D}}}{{\dfrac{{\Delta l}}{l}}}$

**Complete step by step solution:**

Let D and l be the diameter and length of the given wire respectively.

When the wire is stretched, the length increases by 0.05%.

We know, the Poisson’s ratio $\sigma = \dfrac{{{\text{lateral strain}}}}{{{\text{longitudinal strain}}}}$

$ \Rightarrow \sigma = \dfrac{{\dfrac{{\Delta D}}{D}}}{{\dfrac{{\Delta l}}{l}}} = 0.4$

Here, $\dfrac{{\Delta l}}{l}$= 0.05

$ \Rightarrow \sigma = \dfrac{{\dfrac{{\Delta D}}{D}}}{{0.05}} = 0.4$

$ \Rightarrow \dfrac{{\Delta D}}{D} = 0.05 \times 0.4 = 0.02$

**Therefore, the correct answer is option (A), reduced by 0.02%.**

**Note:**If the length of the wire increases on stretching, the diameter will decrease simultaneously. However, the ratio of lateral and longitudinal strain is always a constant for a material and is defined as the Poisson’s Ratio. It is denoted by the letter $\sigma $.

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