Answer
$2Hg(NO_3)_2(s)\rightarrow 2HgO(s)+4NO_2(g)+O_2(g)$
Work Step by Step
Mercury (II) nitrate is $Hg(NO_3)_2(s)$. Mercury (II) oxide is $HgO(s)$. Nitrogen dioxide is $NO_2(g)$. Oxygen is $O_2(g)$
$Hg(NO_3)_2(s)\rightarrow HgO(s)+NO_2(g)+O_2(g)$
Add a coefficient of 2 to $NO_2$ to balance $N$.
$Hg(NO_3)_2(s)\rightarrow HgO(s)+2NO_2(g)+O_2(g)$
There is an odd number of $O$ atoms on the right which can never be balanced against the even number on the left. Add a coefficient of 2 to $HgO$ to eliminate the odd.
$Hg(NO_3)_2(s)\rightarrow 2HgO(s)+2NO_2(g)+O_2(g)$
Add a coefficient of 2 to $Hg(NO_3)_2$ to balance $Hg$.
$2Hg(NO_3)_2(s)\rightarrow 2HgO(s)+2NO_2(g)+O_2(g)$
Increase the coefficient to 4 for $NO_2$ to balance $N$.
$2Hg(NO_3)_2(s)\rightarrow 2HgO(s)+4NO_2(g)+O_2(g)$
