To determine the oxidation number of Mn in \( \text{K}_2\text{MnO}_4 \), the following equation is employed:\[2(+1) + x + 4(-2) = 0,\]where \( x \) represents the oxidation state of Mn.
Upon simplification:\[2 + x - 8 = 0 \implies x = +6.\]For the remaining compounds, the oxidation numbers of Mn are as follows:- In \( \text{MnO}_2 \), Mn = \( +4 \),- In \( \text{Mn}_3\text{O}_4 \), Mn = \( +\frac{8}{3} \) (average),- In \( \text{MnSO}_4 \), Mn = \( +2 \).Consequently, the highest oxidation number is observed in \( \text{K}_2\text{MnO}_4 \). Final Answer:\[\boxed{\text{K}_2\text{MnO}_4}\]