Answer
$f = 310~Hz$
Work Step by Step
Let $L$ be the original length of the spring.
We can find $k_1$:
$\frac{7.0}{17} L~k_1 = k~L$
$k_1 = \frac{17}{7.0}k$
$k_1 = (\frac{17}{7.0})~(8600~N/m)$
$k_1 = 20,886~N/m$
We can find he mass of the block:
$f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$
$f^2 = \frac{k}{4\pi^2 m}$
$m = \frac{k}{4\pi^2 f^2}$
$m = \frac{8600 N/m}{(4\pi^2) (200~Hz)^2}$
$m = 0.005446~kg$
We can find the oscillation frequency:
$f = \frac{1}{2\pi} \sqrt{\frac{k_1}{m}}$
$f = \frac{1}{2\pi} \sqrt{\frac{20,886 ~N/m}{ 0.005446~kg}}$
$f = 310~Hz$
