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Okay, since you have asked this question, I hope you are familiar enough with redox reactions, oxidation number, loss gain of electron etc...
So, the reaction is  KMnO4 + SO2+ H2O=K2SO4+MnSO4+ H2SO4 
Now, we see calculate the oxidation number and change in oxidation number per mole of the given compounds.
We see that in KMnO4, O.N of Mn is changing from 7 to 2 [in MnSO4]
Also, the O.N of SO2 is changing from 4 to 6 [in K2SO4] 

NOTE- the oxidation number of S is 6 in all the compounds in the product side, so instead of disproportionate reaction, it is a simple Oxid-Redux Reaction, and we can take any of the compounds in the product side to compare its O.N with SO2 in reactant's side.

Reduction Half
5 e^{-} +KMnO4 --> MnSo4 [O.N of Mn changes from 7 to 2, thus Reduction]

Oxidation Half
For SO2, O.N is changing= from 4 to 6
SO2 [4] ---> Products [6]  + 2 e^{-}

Now, balancing electrons, by multiplying R-Half by 2 and O-Half by 5 we get :-

2KMnO4 + 5SO2 + H2O --> K2SO4 + 2MnSO4 + 2H2SO4

[NOTE - Here we are balancing S in H2SO4 because we should not bother other products as they are already balanced, and we need to balance S and H]
Now balancing H and O:-
2 KMnO4 + 5O2 + 2H2O --> K2SO4 + 2MnSO4 + 2H2SO4

Hence, balanced, I am sure there will be some queries or doubts, please comment below, I am here to help everyday :-)

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