We know

\(\frac{\text { Force } \times \text { Length }}{\text { Area of cross-section } \times \text { elongation }}=\) Young's Modulus

\(\frac{F \times L}{A \times \Delta L}=Y\) \(\left\{\begin{array}{l}F=2 \times 10 N , A=\pi \times(1 / 2)^2 \times 10^{-6} \,m ^2 \\ L=3 \,m \quad, \Delta L=1 \times 10^{-3} \,m \end{array}\right\}\)

Substituting values

\(\frac{20 \times 3}{\pi \times \frac{1}{4} \times 10^{-6} \times 1 \times 10^{-3}}=Y\)

\(\frac{20 \times 3 \times 4}{3.14 \times 10^{-9}}=Y\)

\(7.48 \times 10^{10} \,Nm ^{-2}=Y\)