6. Let $S=[\sqrt{1}]+[\sqrt{2}]+[\sqrt{3}]+\cdots+[\sqrt{2002}]$, where $[\sqrt{n}]$ denotes the greatest integer not exceeding $\sqrt{n}$. Then the value of $[\sqrt{S}]$ is
6.242.

Let $k^{2} \leqslant n<(k+1)^{2}$, then $k \leqslant \sqrt{n}<k+1$, so
$$
\begin{array}{l}
{[\sqrt{n}]=k .} \\
\because(k+1)^{2}-k^{2}=2 k+1,
\end{array}
$$
$\therefore$ From $k^{2}$ to $(k+1)^{2}$ there are $2 k+1$ numbers, the integer parts of whose square roots are all $k$,
$$
\begin{array}{l}
\therefore\left[\sqrt{k^{2}}\right]+\left[\sqrt{k^{2}+1}\right]+\cdots+\left[\sqrt{\left(k^{2}+1\right)^{2}-1}\right] \\
=k(2 k+1)=2 k^{2}+k . \\
\text { Also, } \because 44^{2}<2002=44^{2}+66<45^{2},
\end{array}
$$
$\therefore$ Between 1 and 2002, there are 44 perfect squares
$$
\begin{array}{l}
1^{2}, 2^{2}, \cdots, 44^{2} . \\
\therefore S=[\sqrt{1}]+[\sqrt{2}]+\cdots+[\sqrt{2002}] \\
\quad=2\left(1^{2}+\cdots+43^{2}\right)+(1+\cdots+43)+44 \times 67 \\
\quad=58762 .
\end{array}
$$

Thus $[\sqrt{S}]=[\sqrt{58762}]=242$.