(a) $$\mathop C\limits^{ + 6} r{O_3}$$ : Since there are three oxygen atoms each with an oxidation number of -2, the total oxidation number from the oxygen atoms is -6. To make the compound neutral, Chromium (Cr) must therefore have an oxidation number of +6.

(b) $$\mathop {F{e_2}}\limits^{ + 3} {O_3}$$ : Here, there are three oxygen atoms, contributing a total oxidation number of -6. To balance this, the total oxidation number for the two iron (Fe) atoms must be +6, meaning each Fe atom has an oxidation number of +3.

(c) $$\mathop {Mn}\limits^{ + 4} {O_2}$$ : With two oxygen atoms, the total oxidation number is -4. Thus, the manganese (Mn) atom must have an oxidation number of +4 to balance this.

(d) $$\mathop {{V_2}}\limits^{ + 5} {O_5}$$ : In this case, there are five oxygen atoms for a total oxidation number of -10. To balance this, the total oxidation number for the two vanadium (V) atoms must be +10, meaning each V atom has an oxidation number of +5.

(e) $$\mathop {C{u_2}}\limits^{ + 1} O$$ : Here, there's one oxygen atom with an oxidation number of -2. To balance this, the total oxidation number for the two copper (Cu) atoms must be +2, meaning each Cu atom has an oxidation number of +1.

So, the order of the oxidation numbers from highest to lowest is :

(a) CrO$_3$ > (d) V$_2$O$_5$ > (c) MnO$_2$ > (b) Fe$_2$O$_3$ > (e) Cu$_2$O.

Therefore, the answer is Option C : (a) > (d) > (c) > (b) > (e).