= Solution
Use <cylindrical coordinates> with $dV=r\,dr\,d\theta\,dz$, $0\leq\theta<2\pi$, $0\leq z\leq h$ and $0\leq r\leq a(1-z/h)$. Set $s=1-z/h$. Integrating first over radius gives the mass per unit height
$$
\frac{dM}{dz}=2\pi\rho_0\left[\frac{a^2s^2}{2}+\frac{a^2s^3}{3}\right]=\pi\rho_0a^2\left(s^2+\frac23s^3\right).
$$
Therefore
$$
\boxed{M=\pi\rho_0a^2h\int_0^1\left(s^2+\frac23s^3\right)ds=\frac12\pi\rho_0a^2h.}
$$
For the first moment about the base, use $z=h(1-s)$:
$$
Md=\pi\rho_0a^2h^2\int_0^1(1-s)\left(s^2+\frac23s^3\right)ds=\frac7{60}\pi\rho_0a^2h^2.
$$
Dividing by the total mass gives
$$
\boxed{d=\frac{7h}{30}.}
$$
The <center of mass of a cone with a power-law radial density> gives an independent decomposition: the uniform-density component has centroid $h/4$, while the component proportional to $r$ has centroid $h/5$ and half as much mass. Their mass-weighted mean is $7h/30$. Density increasing toward the wider outer parts consequently moves the center below the uniform cone's $h/4$.
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