**Answer:**

The diameter is 7.64 mm.

**Explanation:**

**Given that,**

Force = 6640 N

Length = 380 mm

Elongation = 0.50 mm

Modulus of elasticity = 110 GPa

**We need to calculate the area**

**Using formula of stress**

[tex]\sigma=\dfrac{F}{A}[/tex]....(I)

[tex]\sigma=\dfrac{E\Delta l}{l}[/tex]....(II)

From equation (I) and (II)

[tex]\dfrac{F}{A}=\dfrac{E\Delta l}{l}[/tex]

[tex]A=\dfrac{Fl}{E\Delta l}[/tex]

Put the value into the formula

[tex]A=\dfrac{6640\times380\times10^{-3}}{110\times10^{9}\times0.50\times10^{-3}}[/tex]

[tex]A=4.587\times10^{-5}\ m^2[/tex]

**We need to calculate the diameter**

[tex]A=\dfrac{\pi d^2}{4}[/tex]

[tex]4.587\times10^{-5}=\dfrac{\pi d^2}{4}[/tex]

[tex]d=\sqrt{\dfrac{4\times A}{\pi}}[/tex]

[tex]d=\sqrt{\dfrac{4\times4.587\times10^{-5}}{\pi}}[/tex]

[tex]d=7.64\ mm[/tex]

**Hence, The diameter is 7.64 mm.**