Step 1: Work out the oxidation number of chromium.
The compound is $[Cr(NH_3)_4(ONO)Cl]NO_3$, an overall neutral salt. Ammine ligands contribute no charge, nitrito ($ONO^-$) and chloro ($Cl^-$) each contribute $-1$, and the counter ion nitrate also carries $-1$. Setting the whole thing to zero: \[ x + 0 - 1 - 1 - 1 = 0 \implies x = +3 \]
Step 2: Alphabetize the ligand names inside the bracket.
The three kinds of ligands are ammine ($NH_3$), chloro ($Cl^-$), and nitrito-O ($ONO^-$, bound through oxygen). Going alphabetically by the ligand name itself, the order is ammine, then chloro, then nitrito-O.
Step 3: Assemble the complex ion's name.
With four ammine ligands we write tetraammine, followed by chloro, then nitrito-O, then the metal name chromium with its oxidation state in Roman numerals, (III).
Step 4: Attach the counter ion to finish the name.
Since the counter ion is $NO_3^-$, the full name ends in nitrate. \[ \boxed{\text{Tetraamminechloronitrito-O-chromium(III) nitrate}} \]