A cylindrical wire of length L and diameter D is subjected to a...

Longitudinal stress is given by $\dfrac{F}{A}$, where $A = \dfrac{\pi D^2}{4}$.

From the definition of Young's modulus ($Y$), the longitudinal strain is:
$\dfrac{\Delta L}{L} = \dfrac{F}{A Y} = \dfrac{4F}{\pi D^2 Y}$

Poisson's ratio ($\sigma$) is the ratio of lateral strain to longitudinal strain:
$\sigma = \dfrac{\Delta D / D}{\Delta L / L}$

Rearranging for the lateral strain:
$\dfrac{\Delta D}{D} = \sigma \left(\dfrac{\Delta L}{L}\right) = \sigma \left(\dfrac{4F}{\pi D^2 Y}\right)$

Therefore, the magnitude of the decrease in diameter is:
$\Delta D = D \times \dfrac{4F\sigma}{\pi D^2 Y} = \dfrac{4F\sigma}{\pi D Y}$

Answer: $\dfrac{4F\sigma}{\pi D Y}$