#### Answer

$f = 1.6~Hz$

#### Work Step by Step

The mass oscillates about the equilibrium point which is $x = 10~cm$ below the point where the mass was released. At this point, the force from the spring is equal in magnitude to the weight. We can write an expression for $\frac{k}{m}$:
$kx = mg$
$\frac{k}{m} = \frac{g}{x}$
We can find the frequency of the oscillations;
$f = \frac{1}{2\pi}~\sqrt{\frac{k}{m}}$
$f = \frac{1}{2\pi}~\sqrt{\frac{g}{x}}$
$f = \frac{1}{2\pi}~\sqrt{\frac{9.80~m/s^2}{0.10~m}}$
$f = 1.6~Hz$