Answer
$n \geq 4$
Work Step by Step
We have $S_n=a_1-a_2+......+(-1)^{n+1}a_n$
Now, $|S-S_n| \leq |a_{n+1}|$
So, $|a_{n+1} | \lt 0.001 $ or, $|\dfrac{1}{((n+1)+3 \sqrt {n+1})^3}| \lt 0.001$
or, $((n+1)+3 \sqrt {n+1})^3 \gt 1000$
or, $n \geq 4$
Thus, we need $n \geq 4$ terms to estimate the sum of the entire series when an error is less than $0.001$.