The IUPAC name of $K_3\left[Ir{\left(C_2{O}_4\right)}_3\right]$  is

According to IUPAC nomenclature, while naming the name of the complex-counter ion is written first. In the given complex, potassium is the counter ion. As the oxidation state of Ir in +3, so Iridium is written as iridate(III). There are three oxalate ion (ligands), so trioxalato is written in IUPAC name of the complex.

Finally, name of the complex is ${\mathrm{K}}_3\left[\mathrm{Ir}{\left({\mathrm{C}}_2{\mathrm{O}}_4\right)}_3\right]$ becomes Potassium trioxalatoiridate (III).