Given, mass of shell, $m_{s}=20 \mathrm{~g}=0.02 \mathrm{~kg}$
Mass of gun, $m_{g}=100 \mathrm{~kg}$
Speed of shell, $v_{s}=80 \mathrm{~m} / \mathrm{s}$
Let $v_{g}$ be the speed of the recoil of gun, then according to law of conservation of linear momentum.
Total initial momentum = Total final momentum
$$
\begin{array}{ll}
\Rightarrow & 0=m_{s} v_{s}+m_{g} v_{g} \\
\Rightarrow & 0=0.02 \times 80+100 v_{g} \\
\Rightarrow & 0=1.6+100 v_{g}
\end{array}
$$
$$
\begin{aligned}
\Rightarrow \quad v_{g} &=\frac{-1.6}{100} \\
&=-1.6 \times 10^{-2} \mathrm{~m} / \mathrm{s}=-1.6 \mathrm{~cm} / \mathrm{s}
\end{aligned}
$$
Negative sign indicates that gun moves in opposite direction to that of shell.
$\therefore$ Speed of recoil of $g u n=1: 6 \mathrm{~cm} / \mathrm{s}$