Step 1: Approach:
Count the carbons and then look for any pi bond. Pi bonds are the only thing that would lower the hybridisation below sp3.
Step 2: Count:
The formula $\text{HO}(\text{CH}_2)_2\text{CH}(\text{CH}_3)_2$ has 2 carbons in the chain from $(\text{CH}_2)_2$, 1 from CH, and 2 from the two methyl groups, total $2+1+2=5$.
Step 3: Pi Bonds:
The molecule has no C=C, C=O or triple bond, so every carbon is saturated and $sp^3$. The answer is 5, option (D).
Final Answer:
Five sp3 carbons, option (D).
\[ \boxed{\text{(D) } 5} \]