We know that, maximum speed along the circular path is given by
$$
v=\sqrt{\mu r g}
$$
where, $r$ is radius of the circular track
(i) Time taken for path $(A \rightarrow B \rightarrow C)$
$$
\begin{aligned}
&S_l(\text { Path length })=\frac{3}{4}(2 \pi 2 R) \\
&\quad=3 \pi R=3 \pi \times 100=300 \pi \mathrm{m}
\end{aligned}
$$
The maximum speed of car along the circular path $\left(v_1\right)$ is
$$
\begin{aligned}
v_1 &=\sqrt{\mu r g}=\sqrt{\mu 2 R g} \\
&=\sqrt{0.1 \times 2 \times 100 \times 10}=10 \times \sqrt{2} \\
&=14.14 \mathrm{~m} / \mathrm{s} \\
t_1 &=\frac{S_1}{v_1}=\frac{300 \pi}{14.14}=66.6 \mathrm{~s}
\end{aligned}
$$
(ii) Time taken for path $(D \rightarrow E \rightarrow F)$
$$
\text { Path length }=\frac{1}{4}(2 \pi R)=\frac{\pi \times 100}{2}=50 \pi
$$
$$
\begin{aligned}
v_2 &=\sqrt{\mu R g}=\sqrt{0.1 \times R \times g} \\
&=\sqrt{0.1 \times 100 \times 10}=10 \mathrm{~m} / \mathrm{s} \\
t_2 &=\left(\frac{S_2}{v_2}\right)=\frac{50 \pi}{10}=5 \pi \mathrm{s}=15.7 \mathrm{sec} .
\end{aligned}
$$
(iii) Time for path $C \rightarrow D \& F \rightarrow A$
For paths, $C D$ and $F A$,
Path length $=C D+F A$
$$
=R+R=2 R=200 \mathrm{~m}
$$
As path $C D$ and $F A$ can be consider as in straight so car will travel with its maximum speed $=50 \mathrm{~m} / \mathrm{s}$
$$
t_3=\frac{S_3}{v_3}=\frac{200}{50}=4.0 \mathrm{~s} .
$$
$\therefore$ Total time for completing one round
$$
t=t_1+t_2+t_2=66.6+15.7+4.0=86.3 \mathrm{~s}
$$