A cylindrical plastic bottle of negligible mass is filled with 500ml of water. On  slightly pressing it downwards and releasing, it starts executing simple harmonic  motion. If radius of the bottle is 5cm, then its angular frequency of oscillation is  close to ($densityofwater\rho ={10}^{3}kg/m^3$)

The restoring force responsible for executing simple harmonic motion in this case  is the buoyant force.
Using this, we can equate  $F_B=\rho Vg$ to the restoring force $F=kx$ .
Comparing the two equations,  $kx=\rho Vg$
Here  $V=Ax$,
Substituting this we get  $kx=\rho Axg$
Hence,  $k=\rho Ag$ 
For a body executing simple harmonic motion, its angular frequency can be written  as   $\omega =\sqrt{\frac{k}{m}}$
Substituting $k=\rho Ag$ in  $\omega =\sqrt{\frac{k}{m}}$, we get 
$\omega =\sqrt{\frac{\rho Ag}{m}}$ .
Now, the mass of water in the bottle  $m=\rho Ah$
Hence,   $\omega =\sqrt{\frac{\rho Ag}{\rho Ah}}=\sqrt{\frac{g}{h}}$………………………(1)
In the question, the volume of water is given as 500 ml. Using the conversion that  $1l={10}^{-3}{m}^3,$  we get  $1ml={10}^{-6}{m}^3,$
Hence, volume of 500ml of water is   $5\times {10}^{-4}{m}^3$
Again using  $V=Ah$,  $5\times {10}^{-4}=Ah$
Therefore  $h=\frac{5\times {10}^{-4}}{A}=\frac{5\times {10}^{-4}}{\pi r^2}$. Given the radius if the base is 5 cm,
$h=\frac{5\times {10}^{-4}}{\pi \times {\left(5\times {10}^{-2}\right)}^2}$ = $\frac{1}{5\pi }$
$h=\frac{1}{15.70}$ .
Substituting this in equation (1),
 $\omega =\sqrt{\frac{g}{h}}=\sqrt{\frac{9.8}{\frac{1}{15.70}}}$

$\omega =12.7rad/s$