The density of copper at $0^{0}C$ is $9gmc{c}^{-1}.$ If $\alpha$ of copper $=17\times {10}^{-6}{/}^{\circ }C,$ it’s density at ${20}^{0}C$ is

 $s_1=9gm/cc\Rightarrow \alpha =17\times {10}^{-6}{/}^{0}c\Rightarrow 1^{\gamma }=3\alpha$
 $t_1=0^{0}c\Rightarrow t_2={20}^{0}c$
 $s_2=?$
 $s_2=\frac{s_1}{1+\gamma (t_2-t_1)}\Rightarrow s_2=\frac{9}{1+51\times {10}^{-6}\left(20\right)}=\frac{9}{1+1020\times {10}^{-6}}=\frac{9}{1+0.00102}$
 $=\frac{9}{1.00102}=8.99gm/cc$