Answer
$2KClO_3(s)\rightarrow 2KCl(s)+3O_2(g)$
Work Step by Step
$KClO_3(s)\rightarrow KCl(s)+O_2(g)$
$K$ and $Cl$ are balanced, but the left side has an odd number of $O$ atoms which only appear in pairs on the right side. Add a coefficient of 2 to $KClO_3$ to create an even number of $O$ atoms on the left side. This requires you to also add a coefficient of 2 to $KCl$ to keep $K$ and $Cl$ in balance.
$2KClO_3(s)\rightarrow 2KCl(s)+O_2(g)$
The left side now has 6 $O$ atoms, so add a coefficient of 3 to $O_2$.
$2KClO_3(s)\rightarrow 2KCl(s)+3O_2(g)$
