how can we balance following equation by ion electron methodKMnO4 + SO2 = MnSO4 + H2SO4 + K2SO4

the equation is wrong,please correct it it should be
2KMnO4 +5SO2 +2H2O --------2MnSO4 + 2KHSO4 +H2SO4
First, on the right, we have 3 Sulphur atoms, so we know that the SO2 on the left must have a coefficient of at least 3:
KMnO4 + 3SO2 + H2O → MnSO4 + KHSO4 + H2SO4

Also, since there are an even number of Oxygen atoms on the right, and must always be, H2o MUST also have an even coefficient, so start with it at a 2:
KMnO4 + 3SO2 + 2H2O → MnSO4 + KHSO4 + H2SO4

Now, since H2O has an even number of H, that means there needs to be an even number of H on the right. Since KHSO4 has only one H, it will need to have a coefficient of at least 2:
KMnO4 + 3SO2 + 2H2O → MnSO4 + 2KHSO4 + H2SO4

This means that KMnO4 on the left needs a coefficient of 2 as well:
2KMnO4 + 3SO2 + 2H2O → MnSO4 + 2KHSO4 + H2SO4

Thus, the coefficient of MnSO4 on the right must be incresed:
2KMnO4 + 3SO2 + 2H2O → 2MnSO4 + 2KHSO4 + H2SO4

Now return your attention to the S, O, and H. On the left there are16 Oxygen atoms, 3 Sulphur atoms, and 4Hydrogen atoms. On the right we have 20 Oxygen atoms, 5 Sulphur atoms, and 4Hydrogen atoms. Look closely now, the H seems to be balanced too, doesn't it!?
So now we have the Sulphur and Oxygen atoms left to balance. This can easily be fixed by increasing the coefficient on SO2 by two:

2KMnO4 + 5SO2 + 2H2O → 2MnSO4 + 2KHSO4 + H2SO4