Answer
$29.5mm$
Work Step by Step
We can determine the required amplitude as follows:
$K=\frac{F}{\delta}$
$\implies K=\frac{18}{0.014}=17.928rad/s$
Now $x=\frac{F_{\circ/K}}{1-(\frac{\omega_{\circ}}{\omega_n})^2}$
We plug in the known values to obtain:
$x=\frac{0.015}{1-(\frac{2\times 2\pi}{17.98})^2}$
$\implies x=0.0295m=29.5mm$
