Given that the density of the cone is half of the density of the water. If the height of the cone that is immersed in water at equilibrium is $h$, then for floating

$$\begin{gathered} \frac{1}{3}\pi R^{2}H{\rho }_C=\frac{1}{3}\pi r^{2}h{\rho }_W \\ \Rightarrow \frac{r^{2}h}{R^{2}H}=\frac{{\rho }_C}{{\rho }_W}=\frac{1}{2} \\ \Rightarrow \frac{h^3}{H^3}=\frac{1}{2} \\ \Rightarrow h={\left(\frac{1}{2}\right)}^{\frac{1}{3}}H \end{gathered}$$

Now 

$$\begin{gathered} \tan 30°=\frac{r}{h} \\ \Rightarrow r=\frac{h}{\sqrt{3}} \end{gathered}$$

If the cone is displaced by $dh$ downward, extra force due to buoyancy on the cone will be,

$F=\left(\pi r^2{\rho }_{W}g\right)dh$

The mass of the cone is

$M=\frac{1}{3}\pi R^{2}H{\rho }_C$

Therefore, the frequency of the SHM will be,

$$\begin{gathered} f=\frac{1}{2\pi }\sqrt{\frac{K}{M}}=\frac{1}{2\pi }\sqrt{\frac{\pi r^2{\rho }_{W}g}{\frac{1}{3}\pi R^{2}H{\rho }_C}}=\frac{1}{2\pi }\sqrt{\frac{3\times 2g}{H}\times \frac{h^2}{H^2}} \\ \Rightarrow f=\frac{1}{2\pi }\sqrt{\frac{6g}{H}\frac{1}{4^{{\displaystyle \frac{1}{3}}}}} \end{gathered}$$