SO3(2-) + H2O = SO4(2-) + 2H(1+) + 2e(-) oxidation - electrons on the right side
MnO4(1-) + 8H(1+) + 5e(-) = Mn(2+) + 4H2O reduction - electrons on the left side
now we multiply the first equation by 5, and the second one by 2, and add both by sides
5SO3(2-) + 5H2O + 2MnO4(1-) + 16H(1+) + 10e(-) = 5SO4(2-) + 10H(1+) + 10e(-) + 2Mn(2+) + 8H2O
and finally
5SO3(2-) + 2MnO4(1-) + 6H(1+) = 5SO4(2-) + 2Mn(2+) + 3H2O