Let the rod be AB and the weight be hung at 0. Let $AO=x$ and the $OB=2-x$
Let $T_1$ and $T_2$ be tension in the two wires.
By principle of moments
$T_1 \times OA=T_2 \times OB$
$\large\frac{T_1}{T_2}=\frac{OB}{OA}$------(1)
Also it is given that the two wires have same stress.
$\therefore \large\frac{T_1}{A_1}=\frac{T_2}{A_2}$
Where $A_1$ and $A_2$ are area of cross - sections.
$\large\frac{T_1}{T_2}=\frac{A_1}{A_2}$
$\qquad=\large\frac{10^{-3}}{2 \times 10^{-3}}=\frac{1}{2}$ -------(2)
From (1) and (2)
$\large\frac{1}{2} =\frac{2-x}{x}$
$x= 4-2x$
$3x= 4$
$\therefore x= \large\frac{4}{3}$
$\qquad= 1.33\; m$
Hence a is the correct answer.