A block is fastened to a horizontal spring. The block is pulled to a distance x = 10   cm from its…

At first the block is pulled to a distance $x_0=10 \mathrm{cm}$(extreme position).

Potential energy of block is $U_i=\frac{1}{2}k{x}_0^2$ and kinetic energy os block, $K_{\mathit{i}}=0$.

Now, when the block is at distance $5 \mathrm{cm}$.

Potential energy is $U_f=\frac{1}{2}k{\left(\frac{x_0}{2}\right)}^2$ and kinetic energy is $K_{\mathit{f}}=0.25 \mathrm{J}$

Total energy of block will be conserved, so 

$\frac{1}{2}k{x}_0^2+0=\frac{1}{2}k\frac{x_0^2}{4}+0.25$

$\Rightarrow \frac{1}{2}k{x}_0^2\frac{3}{4}=\frac{1}{4}$

$$\begin{gathered} \Rightarrow \frac{1}{2}k\frac{3}{100}=1 \\ \Rightarrow k=\frac{200}{3} \mathrm{N} {\mathrm{m}}^{-1} \end{gathered}$$

$=67 \mathrm{N} {\mathrm{m}}^{-1}$