A thin rod having a length of 1   m and area of cross-section 3 × 10 - 6   m 2 is suspended…

Let $\Delta l$ be the decease in length of rod due to decease in temperature

Then, $\Delta l=l\alpha \Delta T$, here, coefficient of linear expansion is $\alpha =2\times {10}^{-5} {\mathrm{K}}^{-1}$ and $\mathrm{\Delta }T=(483-433) \mathrm{K}$$=50 \mathrm{K}$

Thus, $\Delta l=1\times 2\times {10}^{-5}\times 50={10}^{-3} \mathrm{m}$

Young's modulus is stated as  $Y=\frac{{\displaystyle \frac{{\displaystyle F}}{A}}}{{\displaystyle \frac{\Delta {\displaystyle l}}{l}}}$, here, $F=Mg$.

So, $2\times {10}^{11}=\frac{Mg\times 1}{3\times {10}^{-6}\times {10}^{-3}}$

$\Rightarrow Mg=2\times {10}^{11}\times 3\times {10}^{-9}=6\times {10}^{-2}$

Thus, $M=60 \mathrm{kg}$