Answer
$a_{max}=37.8m/s^2$
Work Step by Step
To find the maximum acceleration, use the relation that $$a_{max}=A\omega ^2$$ for an undamped oscillator. To find omega, substitute the relation $$\omega=2\pi f$$ into the equation to get $$a_{max}=A(2\pi f)^2 = 4\pi^2Af^2$$ Substitute known values of $f=6.60Hz$ and $A=0.0220m$ to get a maximum acceleration of $$a_{max}=4\pi ^2(0.0220m)(6.60Hz)^2=37.8m/s^2$$
