The distance covered by rocket after $20 \mathrm{s}$, $h_1=\frac{1}{2}a{t}^2=\frac{1}{2}\times 1\times {\left(20\right)}^2=200 \mathrm{m}$ and, the velocity of the brake piece from the rocket is, $v=0+at=20 \mathrm{m} {\mathrm{s}}^{-1}$, it means the addition distance covered by piece in the upward direction is, $0=v^2-2g{h}_2\Rightarrow h_2=\frac{v^2}{2g}=\frac{400}{2\times 10}=20 \mathrm{m}$, so the time take by the rocket to cover a distance of $20 \mathrm{m}$, $0=20-10t_2\Rightarrow t_2=2 \mathrm{s}$, and the total hight from the ground becomes, $h_3=h_1+h_2=200+20=220 \mathrm{m}$, so the time of fall becomes, 
$t_3=\sqrt{\frac{2h_3}{g}}=\sqrt{\frac{2\times 220}{10}}=2\sqrt{11} \mathrm{s}$ and net time to reach at ground, $2\sqrt{11} \mathrm{s}+2 \mathrm{s}=8.6 \mathrm{s}$