The cross-sectional area of the lead bar is [tex]9.188 \times 10^{-4} \ m^2[/tex].
The Young's modulus (E)of lead bar is given as 1.6 x 10¹⁰ Pa.
The cross-sectional area of the bar is calculated as follows;
[tex]E = \frac{stress}{strain} = \frac{F/A}{e/l} = \frac{Fl}{Ae} \\\\A = \frac{Fl}{Ee}[/tex]
where;
[tex]A = \frac{(mg) l}{Ee} \\\\A = \frac{(10^3 \times 9.8) \times 3}{1.6 \times 10^{10} \times 2 \times 10^{-3}} \\\\A = 9.188 \times 10^{-4} \ m^2[/tex]
Thus, the cross-sectional area of the lead bar is [tex]9.188 \times 10^{-4} \ m^2[/tex].
Learn more about Young's modulus here: https://brainly.com/question/6864866
