II. (40 points) Find the smallest real number $a$ such that for all positive integers $n$ and real numbers $0=x_{0}<x_{1}<\cdots<x_{n}$, the following inequality holds:
$$
a \sum_{k=1}^{n} \frac{\sqrt{(k+1)^{3}}}{\sqrt{x_{k}^{2}-x_{k-1}^{2}}} \geqslant \sum_{k=1}^{n} \frac{k^{2}+3 k+3}{x_{k}} .
$$
$$
\frac{(k-1)^{3}}{x_{k-1}}+\frac{(8 k+8)^{\frac{3}{2}}}{\sqrt{x_{k}^{2}-x_{k-1}^{2}}}>\frac{(k+3)^{3}}{x_{k}} \text {. }
$$

The above inequality holds for $k=2,3, \cdots, n$. Summing the inequality over $k=2,3, \cdots, n$, and noting that
$$
\frac{(8 \times 1+8)^{\frac{3}{2}}}{\sqrt{x_{1}^{2}-x_{0}^{2}}}=\frac{(1+3)^{3}}{x_{1}},
$$

we get
$$
\begin{array}{l}
\sum_{k=1}^{n} \frac{(8 k+8)^{\frac{3}{2}}}{\sqrt{x_{k}^{2}-x_{k-1}^{2}}} \\
\geqslant \sum_{k=1}^{n} \frac{(k+3)^{3}-k^{3}}{x_{k}}+\frac{n^{3}}{x_{n}} \\
\geqslant \sum_{k=1}^{n} \frac{9\left(k^{2}+3 k+3\right)}{x_{k}} .
\end{array}
$$

When $a=\frac{16 \sqrt{2}}{9}$, the original inequality holds.
Additionally, taking $x_{n}=n(n+1)(n+2)(n+3)$, where $x_{k-1}=\frac{k-1}{k+3} x_{k}(k=2,3, \cdots, n)$, as $n \rightarrow \infty$, $a$ can be achieved.