Given that:

Length of the steel wire  = 1.0 m

Area of cross-section, A = $0.50\times {10}^{-2} {\mathrm{cm}}^2=0.50\times {10}^{-6} {\mathrm{m}}^2$

A mass 100g is suspended from its midpoint.

The wire dips at the centre as shown in the given figure. 

The length after mass m is attached to the wire = XO + OZ

Where,

$\mathrm{XO}=\mathrm{OZ}={\left[{\left(0.5\right)}^2+{\mathrm{l}}^2\right]}^{\frac{1}{2}}$
$\mathrm{Increase} \mathrm{in} \mathrm{the} \mathrm{length} \mathrm{of} \mathrm{the} \mathrm{wire}:$
$\mathrm{\Delta l} = (\mathrm{XO}+\mathrm{OZ})–\mathrm{XZ} = (\mathrm{XO}+\mathrm{OZ})-1.0$
$∆\mathrm{l}=2{\left[{\left(0.5\right)}^2+{\left(\mathrm{l}\right)}^2\right]}^{\frac{1}{2}}-1.0$
$$

Expanding and neglecting higher terms,

$$
$∆\mathrm{l}=\frac{{\mathrm{l}}^2}{0.5}$
$\mathrm{Strain}=\frac{\mathrm{lncrease} \mathrm{in} \mathrm{length}}{\mathrm{Orignal} \mathrm{length}}$

If T is the tension in the wire,

mg = 2Tcosθ

From the figure,

$\mathrm{cos\theta }=\frac{\mathrm{l}}{(0.5){\left(1+{\left({\displaystyle \frac{\mathrm{l}}{0.5}}\right)}^2\right)}^{{\displaystyle \frac{1}{2}}}}$

Expanding the expression and eliminating the higher terms:

$\mathrm{cos\theta }=\frac{\mathrm{l}}{(0.5)\left(1+{\displaystyle \frac{{\mathrm{l}}^2}{2{(0.5)}^2}}\right)}$
$\left(1+\frac{{\mathrm{l}}^2}{0.5}\right)\simeq 1 \mathrm{for} \mathrm{small} \mathrm{l}$
$\Rightarrow \mathrm{cos\theta }=\frac{\mathrm{l}}{0.5}$
$\Rightarrow \mathrm{T}=\frac{\mathrm{mg}}{2\left({\displaystyle \frac{\mathrm{l}}{0.5}}\right)}=\frac{\mathrm{mg}\times 0.5}{2\mathrm{l}}=\frac{\mathrm{mg}}{4\mathrm{l}}$
$\mathrm{Young}\text{'}\mathrm{s} \mathrm{modulus}=\frac{\mathrm{Stress}}{\mathrm{Strain}}$
$\mathrm{Y}=\frac{\mathrm{mg}\times 0.5}{4\mathrm{l}\times \mathrm{A}\times {\mathrm{l}}^2}$
$\mathrm{l}=\sqrt[3]{\frac{\mathrm{mg}\times 0.5}{4\mathrm{YA}}}$

Young’s modulus of steel, Y = $2\times {10}^{11} \mathrm{Pa}$

$\therefore \mathrm{l}=\sqrt{\frac{0.1\times 9.8\times 0.5}{4\times 2\times {10}^{11}\times 0.50\times {10}^{-6}}}$
$=0.0106 \mathrm{m}$

Hence, the depression at the midpoint is 0.0106 m.