As shown in the figure, forces of 105N each are applied in opposite directions, on the upper and lower faces of a cube of side 10 cm, shifting the  upper face parallel to itself by 0.5 cm. If the side of another cube of the same material is 20 cm, then under similar conditions as above, the displacement will be 

Given,

force $\mathrm{F}={10}^5\mathrm{N}$

Area $\mathrm{A}=0.1{\mathrm{m}}^2$

change in length = 0.5 cm

We know that,

$\mathrm{stress}=\frac{\mathrm{force}}{\mathrm{area}} \mathrm{and} \mathrm{Strain}=\frac{∆\mathrm{l}}{\mathrm{l}}$

${\mathrm{Stress}}_1=\frac{{10}^5}{{(0.1)}^2}$

${\mathrm{Strain}}_1=\frac{0.5\times {10}^{-2}}{0.1}$

Consider calculating the stress and strain for block 2 in the second block.
Here, X represents the second block's displacement.

Area $\mathrm{A}=0.2{\mathrm{m}}^2$

${\mathrm{Stress}}_1=\frac{{10}^5}{{(0.2)}^2}$

${\mathrm{Strain}}_1=\frac{\mathrm{X}}{0.2}$

taking the ratio,

$\frac{{\mathrm{stress}}_1}{{\mathrm{strain}}_1}=\frac{{\mathrm{stress}}_2}{{\mathrm{strain}}_2}$

$\frac{{\displaystyle \frac{{10}^5}{0.1^2}}}{{\displaystyle \frac{0.5\times {10}^{-2}}{0.1}}}=\frac{{\displaystyle \frac{{10}^5}{0.2^2}}}{{\displaystyle \frac{\mathrm{X}}{0.2}}}$

$\mathrm{x}=0.25\mathrm{cm}$

Hence the correct answer is 0.25 cm.