a
We know that
$E =\frac{\frac{ F }{ A }}{\frac{\Delta L }{ L }} \Rightarrow \frac{\Delta L }{ L }=\frac{ F }{ AE } \ldots . . \text {. (i) }$
$\text { Also }$
$\frac{\Delta L }{ L } \propto \alpha \Delta T . \ldots . \text { (ii) }$
$\text { from (i) and (ii) }$
$\frac{ F }{ AE }=\alpha \Delta T$
$\Rightarrow mg =(a \Delta T ) AE$
$\Rightarrow m =g a \Delta TAE$
$=\frac{10^{-5} \times 10 \times \pi \times 10^{-6} \times 10^{11}}{10}$
$=\pi \approx 3$
So correct answer is a