Answer
$2Fe(OH)_3(s)+3H_2SO_4(s)\rightarrow Fe_2(SO_4)_3(aq)+6H_2O(l)$
Work Step by Step
$Fe(OH)_3(s)+H_2SO_4(s)\rightarrow Fe_2(SO_4)_3(aq)+H_2O(l)$
Add a coefficient of 2 to $Fe(OH)_3(s)$ to balance $Fe$. Add a coefficient of 3 to $H_2SO_4$ to balance S.
$2Fe(OH)_3(s)+3H_2SO_4(s)\rightarrow Fe_2(SO_4)_3(aq)+H_2O(l)$
There are now 18 $O$ atoms on the left and 13 on the right so an additional 5 are needed. There are now 12 $H$ atoms on the left and 2 on the right, so an additional 10 are needed. Increase the coefficient on $H_2O$ from 1 to 6 to add the necessary atoms.
$2Fe(OH)_3(s)+3H_2SO_4(s)\rightarrow Fe_2(SO_4)_3(aq)+6H_2O(l)$
