Answer
It would take 0.233 seconds to reach the equilibrium position again.
Work Step by Step
We can find the spring constant $k$ as:
$kx = mg$
$k = \frac{mg}{x}$
$k = \frac{(1.65~kg)(9.80~m/s^2)}{0.215~m}$
$k = 75.2~N/m$
We can find the period of oscillations of this mass on the spring as:
$T = 2\pi~\sqrt{\frac{m}{k}}$
$T = 2\pi~\sqrt{\frac{1.65~kg}{75.2~N/m}}$
$T = 0.9307~s$
To reach the equilibrium position again, the mass needs to move through one-quarter of a cycle, which would require a time of one-quarter of the period. We can find the required time $t$.
$t = \frac{T}{4}$
$t = \frac{0.9307~s}{4}$
$t = 0.233~s$
It would take 0.233 seconds to reach the equilibrium position again.
