Answer
$t = 14.3 \,\text{s}$
Work Step by Step
We have two equations
\begin{equation}
x^o(t)=x_{m} \cos \left(\omega^{\prime} t+\phi\right)
\end{equation}
and
\begin{equation}
x(t)=x_{m} e^{-b t / 2 m} \cos \left(\omega^{\prime} t+\phi\right)
\end{equation}
$x^o(t)$ reduces $1/3$ of its value. So, we could get the next
\begin{gather*}
x(t)= \dfrac{1}{3} \, x^o(t) \\
x_{m} e^{-b t / 2 m} \cos \left(\omega^{\prime} t+\phi\right) = \dfrac{1}{3} \, x_{m} \cos \left(\omega^{\prime} t+\phi\right) \\
e^{-b t / 2 m} = \dfrac{1}{3} \\
\dfrac{-bt}{2m} = \ln (1/3) \\
t = \dfrac{- 2 \ln (1/3) m }{b}\\
\end{gather*}
Subsitute with the values of $m$ and $b$
\begin{align*}
t &= \dfrac{- 2 \ln (1/3) m }{b}\\
&= \dfrac{- 2 \ln (1/3) (1.5 \,\text{kg}) }{(0.230 \,\text{kg/s})}\\
&= \boxed{14.3 \,\text{s}}
\end{align*}
