Step 1: Write the full formula:
$\text{CH}_3-\text{CHBr}-\text{CH(CH}_3)-\text{CH(CH}_3)-\text{CH(CH}_3)-\text{CH}_3$.
Step 2: Test each carbon:
C1 and C6 are $\text{CH}_3$ groups, so they cannot be chiral.
C5 is bonded to two $\text{CH}_3$ groups (the substituent and C6), so it has two same groups.
C2, C3 and C4 each have four different groups around them, so these three are stereocentres.
Final Answer:
The number of chiral carbons is $3$, option (C).
\[ \boxed{3} \]