Hilbert class field of $\mathbb{Q}(\sqrt{730})$.

For finding unramified cubic extensions of ${\mathbb Q}(\sqrt{m}\,)$, you have to find singular primary numbers (cubes of ideals) in ${\mathbb Q}(\sqrt{-3m})$. A quick search shows you that $\alpha = 197 + 3\sqrt{-3 \cdot 730}$ is such a number since its norm is $49^3$ and it is congruent to an integer modulo $3 \sqrt{-3}$ (read this with a grain of salt).

Now set $\theta = \sqrt[3]{\alpha} + \sqrt[3]{\alpha'}$; then \begin{align*} \theta^3 & = (\sqrt[3]{\alpha} + \sqrt[3]{\alpha'})^3 = \alpha + 3\sqrt[3]{\alpha}^2\sqrt[3]{\alpha'} + 3\sqrt[3]{\alpha}\sqrt[3]{\alpha'}^2 + \alpha' \\ & = (\alpha + \alpha') + 3\sqrt[3]{\alpha \alpha'}(\sqrt[3]{\alpha} + \sqrt[3]{\alpha'}) = 2 \cdot 197 + 3 \cdot 49 \cdot \theta, \end{align*} hence $\theta$ is a root of the polynomial $f(x) = x^3 - 147 x - 394$. pari confirms that adjoining a root of $f$ to ${\mathbb Q}(\sqrt{730}\,)$ gives you an unramified cyclic extension. More details concerning the construction can be found here.