#### Answer

$g = 5.87~m/s^2$

#### Work Step by Step

We can find the period of the oscillations as:
$T = \frac{time}{oscillations}$
$T = \frac{14.5~s}{10~oscillations}$
$T = 1.45~s$
We then use the period to find the spring constant:
$T= 2\pi~\sqrt{\frac{m}{k}}$
$k = \frac{(2\pi)^2~m}{T^2}$
$k = \frac{(2\pi)^2~(0.200~kg)}{(1.45~s)^2}$
$k = 3.76~N/m$
When the spring is stretched 31.2 cm, the force of the spring is equal to the weight of the mass. We can find $g$ as:
$mg = kx$
$g = \frac{kx}{m}$
$g = \frac{(3.76~N/m)(0.312~m)}{0.200~kg}$
$g = 5.87~m/s^2$