At the highest position;
$\mathit{h}=5\mathrm{m};\mathit{m}=10\mathrm{kg};\mathit{g}=10\mathrm{m}∕{\mathrm{}\mathit{s}}^2$

$=10\times 10\times 5$
$=500\mathrm{J}$
Kinetic Energy $\left({\mathit{K}}_1\right)=0$
$\therefore$ Total mechanical Energy $={\mathrm{U}}_1+{\mathrm{K}}_1$
$=500+0$
$=500\mathrm{J}$
At the height of $2\mathrm{m}$, the total mechanical energy will remain same as it was at the highest position. ( $\because \mathrm{U}+\mathrm{K}=$ constant, for a freely falling body)
$\therefore$ Total mechanical energy $=500\mathrm{J}$