PbO2 ------- PbO dG(298k)

SnO2 ------- SnO dG(298k) 0

Most probable oxidation state of Pb and Sn will be

1- Pb4+ , Sn4+ 2- Pb4+,Sn2+

3- Pb2+, Sn2+ 4- Pb2+, Sn4+

PbO₂ → PbO    ∆G(298k) <0
SnO₂  → SnO   ∆G(298k) > 0

The most probable oxidation state of Pb and Sn would be +2 and +4. The option 4 is correct.
In case of PbO₂ , free energy change is -ve, that is it is favourable for forward reaction. The oxidation state of Pb in PbO is +2.
In case of SnO₂, free energy change is +ve that is forward reaction is not favoured. The oxidation state of Sn in SnO₂ is +4.