A cleaner route is to first find the moment of inertia about the base of the triangle, since that value is well known, and then shift again to the vertex.
About the base, parallel to it, the standard result is \(I_{base} = \dfrac{bh^3}{12}\). This can also be checked using the parallel axis theorem from the centroid, \(I_{base} = \dfrac{bh^3}{36} + \dfrac{1}{2}bh\left(\dfrac{h}{3}\right)^2 = \dfrac{bh^3}{36} + \dfrac{bh^3}{18} = \dfrac{3bh^3}{36} = \dfrac{bh^3}{12}\), which matches.
Now the vertex is at height h, so its distance from the centroid is \(\dfrac{2h}{3}\), while the base's distance from the centroid is \(\dfrac{h}{3}\). Using the parallel axis theorem directly from the centroid for the vertex axis, \(I_{vertex} = \dfrac{bh^3}{36} + \dfrac{1}{2}bh\left(\dfrac{2h}{3}\right)^2 = \dfrac{bh^3}{36} + \dfrac{2bh^3}{9} = \dfrac{bh^3}{4}\).
Taking the ratio, \(\dfrac{I_{vertex}}{I_{cg}} = \dfrac{bh^3/4}{bh^3/36} = 9\).
\[\boxed{9\ \text{times}}\]