Due to momentum of photon, nucleus possesses recoil momentum,

$\because E=m{c}^2$

$=\left(mc\right)c$ $\left[p=mc\right]$

$=pc$

$\therefore$ Momentum of $\gamma$ -photon

$p_1=\frac{E}{c}=p$ ......(i)

According to the law of conservation of linear momentum

$p_i=p_f$

$O=p_1+p_2$

Where, $p_2$ is linear momentum of nucleus

$\Rightarrow p_2=-p_1$

$=\frac{E}{c}$ [in magnitude] .....(ii)

$\because$ Kinetic energy $=\frac{1}{2}m{v}^2$ [From equation (i)]

$=\frac{1}{2}\frac{{\left(mv\right)}^2}{m}$

$\therefore p=mv$

$=\frac{1}{2}\frac{{\left(\frac{E}{c}\right)}^2}{m}$ [from equation (ii)]

$\therefore KE=\frac{1}{2}\frac{{\left(6\times 10^6\times 1.6\times {10}^{-19}/3\times {10}^8\right)}^2}{20\times 1.6\times 10^{-27}}$

$=0.16\times 10^{-15}J$

$=\frac{1.6\times 10^{-16}}{1.6\times 10^{-16}}\mathrm{ k}\mathrm{e}\mathrm{V}$

$=1 \mathrm{ k}\mathrm{e}\mathrm{V}$