Step 1: Understand the goal.
We must pick the most stable carbon free radical from four alkyl radicals.
Step 2: Recall what stabilises a radical.
A free radical has an electron-deficient carbon. Anything that pushes electron density toward that carbon stabilises it - mainly the $+I$ effect of attached alkyl groups and hyperconjugation from neighbouring C-H bonds.
Step 3: Classify each radical.
$\text{CH}_3^{\bullet}$ is methyl (no alkyl groups), $\text{CH}_3\text{CH}_2^{\bullet}$ is primary, $(\text{CH}_3)_2\text{CH}^{\bullet}$ is secondary, and $(\text{CH}_3)_3\text{C}^{\bullet}$ is tertiary.
Step 4: Count electron-donating neighbours.
More alkyl groups attached to the radical carbon means more $+I$ donation and more hyperconjugating C-H bonds. The tertiary radical has three methyl groups feeding it, the maximum here.
Step 5: Apply the stability order.
Stability follows $3^{\circ} > 2^{\circ} > 1^{\circ} > \text{methyl}$, exactly the trend seen for carbocations because both are electron deficient.
Step 6: Conclude.
The tertiary $(\text{CH}_3)_3\text{C}^{\bullet}$ radical sits at the top of this order, so it is the most stable. This is option (2).
\[ \boxed{(\text{CH}_3)_3\text{C}^{\bullet} \text{ is the most stable radical}} \]