Answer
$t= 17 \times 10^{-4} \,\text{s}$
Work Step by Step
The position $+0.800 x_m$ is the origin position at period $t = 0$. So, we find $\phi$ by
\begin{gather*}
x=x_{m} \cos (\omega t+\phi) \\
0.8 x_{m} =x_{m} \cos (0+ \phi) \\
\phi =\cos ^{-1}(0.8)\\
\phi = 0.64\, \mathrm{rad}
\end{gather*}
And $\omega$ is calcauted by
\begin{align*}
\omega &=\sqrt{\frac{k}{m}} \\
&=\sqrt{\frac{1500 \mathrm{N} / \mathrm{m}}{0.055 \mathrm{~kg}}} \\
&=165 \mathrm{~rad/s}
\end{align*}
Sub with $\phi$ and $x = 0.6 x_{m}$ to get the value of $t$
\begin{gather*}
x=x_{m} \cos (\omega t+\phi) \\
0.6 x_{m}=x_{m} \cos ((165 \mathrm{rad/s} ) t + 0.64 \mathrm{rad} )\\
\cos ^{-1}(0.6) = (165 \mathrm{rad/s} ) t + 0.64 \mathrm{rad} \\
t = 17 \times 10^{-4} \,\text{s}
\end{gather*}
