Step 1: Recall stability order of alkyl free radicals.
Stability of free radicals increases with increase in alkyl substitution due to hyperconjugation and +I effect.
\[
\text{Tertiary}>\text{Secondary}>\text{Primary}>\text{Methyl}
\] Step 2: Analyze the given radicals.
R\(_3\)–C\(\cdot\) is tertiary (most stable).
R\(_2\)–CH\(\cdot\) is secondary.
R–CH\(_2\)\(\cdot\) is primary.
CH\(_3\)\(\cdot\) is methyl (least substituted). Step 3: Conclusion.
CH\(_3\)\(\cdot\) has no alkyl group for stabilization, hence it is the most unstable.