#### Answer

$f = 1.7~Hz$

#### Work Step by Step

When the spring is stretched a distance of $x = 9.0~cm$, 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 as:
$f = \frac{1}{2\pi}~\sqrt{\frac{k}{m}}$
$f = \frac{1}{2\pi}~\sqrt{\frac{g}{x}}$
$f = \frac{1}{2\pi}~\sqrt{\frac{9.80~m/s^2}{0.090~m}}$
$f = 1.7~Hz$