We will determine the oxidation state of Mn in each compound:
(1) MnO\(_2\):
Let the oxidation state of Mn be \( x \): \[ x + 2(-2) = 0 \Rightarrow x = +4 \]
(2) MnO\(_4^-\):
\[ x + 4(-2) = -1 \Rightarrow x = +7 \]
(3) KMnO\(_4\):
Potassium is +1, and each oxygen is -2: \[ +1 + x + 4(-2) = 0 \Rightarrow x = +7 \]
(4) Mn\(_2\)O\(_3\):
\[ 2x + 3(-2) = 0 \Rightarrow 2x = +6 \Rightarrow x = +3 \]
The highest oxidation state is \( +7 \), found in both MnO\(_4^-\) and KMnO\(_4\). Therefore, KMnO\(_4\) is the correct compound from the given choices.