Answer
$B_2S_3(s) + 6H_2O(l)\rightarrow 2H_3BO_3(aq)+ 3H_2S(g)$
Work Step by Step
Boron sulfide is $B_2S_3(s)$, water is $H_2O$, boric acid is $H_3BO_3$ and hydrogen sulfide gas is $H_2S$.
$B_2S_3(s) + H_2O(l)\rightarrow H_3BO_3(aq)+ H_2S(g)$
Add a coefficient of 2 to $H_3BO_3$ to balance $B$.
Add a coefficient of 3 to $H_2S$ to balance $S$.
$B_2S_3(s) + H_2O(l)\rightarrow 2H_3BO_3(aq)+ 3H_2S(g)$
Add a coefficient of 6 to $H_2O$ to balane $H$ and $O$.
$B_2S_3(s) + 6H_2O(l)\rightarrow 2H_3BO_3(aq)+ 3H_2S(g)$
