At the lowest position Let T be the tension in the wire.
$T-mg=\large\frac{mv^2}{r}$
$\qquad= mw^2r$----------(1)
Also $Y=\large\frac{T/A}{\Delta l/l}$-----------(2)
$T=mw^2r+mg$
$\quad=m(w^2r +g)$
$\quad=3 \bigg( \large\frac{2 \pi}{2}\bigg)^2$$ \times 1.5 +10$
$T=4.5 \pi ^2+30$
From (2)
$2 \times 10^{11}= \large\frac{3 \pi ^2 +30}{\Large\frac{\pi (1 \times 10^{-3})^2}{\Delta l/1.5}}$
Solving we get
$\Delta l= 7.17 \times 10^{-3}\;m$
Hence b is the correct answer.