Step-by-step Solution:

\[ \textbf{Given: }S_n=630,\quad \text{rows form an AP with common difference }d=-3. \] \[ \text{For }n\text{ rows, let the first row have }a\text{ children.} \] \[ S_n=\frac{n}{2}\bigl(2a+(n-1)(-3)\bigr)=\frac{n}{2}\bigl(2a-3(n-1)\bigr)=630. \] \[ \Rightarrow\;2a-3(n-1)=\frac{1260}{n}\quad\Rightarrow\quad a=\frac{630}{n}+\frac{3}{2}(n-1). \] We test the given options by checking whether \(a\) is a positive integer. \paragraph{Option A: } \(n=5\) \[ a=\frac{630}{5}+\frac{3}{2}(5-1)=126+ \frac{3}{2}\cdot4=126+6=132\quad(\text{integer, valid}). \] \paragraph{Option B: } \(n=4\) \[ a=\frac{630}{4}+\frac{3}{2}(4-1)=157.5+\frac{3}{2}\cdot3=157.5+4.5=162\quad(\text{integer, valid}). \] \paragraph{Option C: } \(n=3\) \[ a=\frac{630}{3}+\frac{3}{2}(3-1)=210+\frac{3}{2}\cdot2=210+3=213\quad(\text{integer, valid}). \] \paragraph{Option D: } \(n=6\) \[ a=\frac{630}{6}+\frac{3}{2}(6-1)=105+\frac{3}{2}\cdot5=105+7.5=112.5\quad(\text{not an integer}). \] Thus \(n=6\) gives a non-integer number of children in the first row, so an arrangement with 6 rows is not possible. \[ \boxed{\text{Answer: D. }6} \]