As given that, $t_1=39.6 \mathrm{~s}, t_2=39.9 \mathrm{~s}$ and $t_3=39.5 \mathrm{~s}$.
The least count of measuring instrument $=0.1 \mathrm{sec}$.
Precision in the measurement is equal to least count of the measuring instrument $(\mathrm{LC})=0.1 \mathrm{sec}$.
Mean value of time for 20 oscillations is
$$
\begin{aligned}
&\bar{t}=\frac{t_1+t_2+t_3}{3} \\
&\bar{t}=\frac{39.6+39.9+39.5}{3}=39.7 \mathrm{sec}
\end{aligned}
$$
then absolute errors in the measurements are :
$$
\begin{aligned}
&\left|\Delta t_1\right|=\left|\bar{t}-t_1\right|=|39.7-39.6|=|0.1 \mathrm{~s}|=0.1 \mathrm{~s} \\
&\left|\Delta t_2\right|=\left|\bar{t}-t_2\right|=|39.7-39.9|=|-0.2 \mathrm{~s}|=0.2 \mathrm{~s} \\
&\left|\Delta t_3\right|=\left|\bar{t}-t_3\right|=|39.7-39.5|=0.2 \mathrm{~s} \mid=0.2 \mathrm{~s}
\end{aligned}
$$
Thus, Mean absolute error
$$
\begin{aligned}
&=\frac{\left|\Delta t_1\right|+\left|\Delta t_2\right|+\left|\Delta t_3\right|}{3} \\
&=\frac{0.1+0.2+0.2}{3}=\frac{0.5}{3}=0.17 \approx 0.2
\end{aligned}
$$
So, accuracy of measurement $=\pm 0.2 \mathrm{~s}$.