$3^{o}$ alcohol $+$ Lucas reagent $\rightarrow$ Immediate turbidity.
$2^{o}$ alcohol $+$ Lucas reagent $\rightarrow$ Turbidity after $5$ mins.
$1^{o}$ alcohol $+$ Lucas reagent $\rightarrow$ No reaction. Thus, the required alcohol is $2 -$ methylpropan $- 2 -$ ol, i.e.
$\mathop {\begin{array}{*{20}{c}}
{\begin{array}{*{20}{c}}
{\,\,\,\,\,\,C{H_3}} \\
| \\
{{H_3}C - C - C{H_3}}
\end{array}} \\
| \\
{\,\,\,\,\,OH}
\end{array}}\limits_{({3^o}\,\,alcohol)} $