Answer
$t = 0.23~ms$
Work Step by Step
The amplitude of the motion is $6.0~nm$
Note that $2.0~nm$ is $\frac{1}{3}$ of the amplitude.
We can find $\theta_1$ when $cos~\theta_1 = \frac{1}{3}$
$\theta_1 = cos^{-1}~\frac{1}{3} = 1.23~rad$
We can find $\theta_2$ when $cos~\theta_2 = -\frac{1}{3}$
$\theta_2 = cos^{-1}~(-\frac{1}{3}) = 1.91~rad$
We can find the phase difference between these two positions:
$1.91~rad-1.23~rad = 0.68~rad$
We can find the time required to move between the two displacements:
$t = \frac{0.68~rad}{2\pi~rad}~T$
$t = (\frac{0.68~rad}{2\pi~rad})~(\frac{2\pi}{\omega})$
$t = (\frac{0.68~rad}{2\pi~rad})~(\frac{2\pi}{3000~rad/s})$
$t = 0.00023~s$
$t = 0.23~ms$
