Correct Answer - Option 2 : 0.033 mm
\({\rm{\Delta }}l\; = \;\frac{F}{A}.\frac{l}{Y}\)
For same F, l, Y
\({\rm{\Delta }}l \propto \frac{1}{A}\)
\(\begin{array}{l}
\therefore \frac{{{\rm{\Delta }}{l_2}}}{{{\rm{\Delta }}{l_1}}} = \frac{{{A_1}}}{{{A_2}}} = \frac{{2\;m{m^2}}}{{6\;m{m^2}}}\\
\therefore {\rm{\Delta }}{l_2} = \frac{1}{3} \times 0.1\;mm = 0.033\;mm
\end{array}\)