Answer
The number of oscillations is 5.3
Work Step by Step
The angular frequency of the damped oscillator $\omega^{\prime}$ is
\begin{align*}
\omega^{\prime}&=\sqrt{\frac{k}{m}-\frac{b^{2}}{4 m^{2}}}\\
& =\sqrt{\frac{8 \mathrm{N} / \mathrm{m}}{1.5 \mathrm{kg}}-\frac{10.23 \mathrm{kg} / \mathrm{s}^{2}}{4(1.50 \mathrm{kg}^2) }}\\
& =2.31 \mathrm{rad} / \mathrm{s}
\end{align*}
Using this value to get the period $T$ by
\begin{align*}
T = \dfrac{2\pi}{\omega^{\prime}} = \dfrac{2\pi}{2.31 \mathrm{rad} / \mathrm{s} } = 2.72 \,\text{s}
\end{align*}
The number of oscillations $n$ equals $(t/T)$
$$ n= \dfrac{t}{ T}= \dfrac{14.3 \mathrm{s}}{ 2.72 \mathrm{s}} = \boxed{5.3} $$
