What is the oxidation state of Vanadium in this solution?

Be careful, since there are two vanadium ions written as VO2+. One of them is (VO2)+, or vanadyl(V), and the other is VO(2+), which is vanadyl(IV). Vanadyl(V) is yellow and vanadyl(IV) is blue, so I guess you meant (VO2)+. THe oxidation state of vanadium plus the oxidation states of the two oxygen atoms (2 � -2 = -4) must equal the charge of the ion (+1), so the vanadium atom is in its +5 oxidation state.