3. Arrange the numbers $\sqrt{2}, 2, \sqrt{6}, 2 \sqrt{2}, \sqrt{10}, \cdots, 2 \sqrt{51}$ as shown in Figure 2.
\begin{tabular}{cccccc}
$\sqrt{2}$ & 2 & $\sqrt{6}$ & $2 \sqrt{2}$ & $\sqrt{10}$ & $2 \sqrt{3}$ \\
$\sqrt{14}$ & 4 & $3 \sqrt{2}$ & $2 \sqrt{5}$ & $\sqrt{22}$ & $2 \sqrt{6}$ \\
$\sqrt{26}$ & $2 \sqrt{7}$ & $\cdots$ & $\cdots$ & $\cdots$ & $\cdots$ \\
$\vdots$ & $\vdots$ & $\vdots$ & $\vdots$ & $\vdots$ & $\vdots$ \\
$\cdots$ & $\cdots$ & $\cdots$ & $\cdots$ & $\cdots$ & $2 \sqrt{51}$
\end{tabular}

If the position of $3 \sqrt{2}$ is denoted as $(2,3)$, and the position of $2 \sqrt{7}$ is denoted as $(3,2)$, then the position of the largest rational number $n$ in this sequence is ( ).
(A) $(16,3)$
(B) $(17,3)$
(C) $(16,2)$
(D) $(17,2)$
3. D.

Observing, we see that this sequence of numbers is $\sqrt{2 k}(k=1,2, \cdots, 102)$, where the largest rational number is $\sqrt{196}$, which is ranked 98th. Therefore, the largest rational number in this sequence is located in the 17th row and the 2nd column.