1. Use the standard form of a travelling wave

If a wave propagates in the positive $x$-direction without changing shape, then its form can be written as

$y(x,t)=f(x-vt)$

where $v$ is the wave speed.

2. Identify the shape at $t=0$

At $t=0$,

$y(x,0)=\frac{1}{(1+x)^2}$

So we can take

$f(x)=\frac{1}{(1+x)^2}$

Hence,

$y(x,t)=\frac{1}{(1+x-vt)^2}$

3. Use the given expression at $t=1\,\text{s}$

At $t=1$,

$y(x,1)=\frac{1}{(1+x-v)^2}$

But the question gives

$y(x,1)=\frac{1}{1+(x-2)^2}$

The printed forms are not of exactly the same shape if read literally. Since the problem states explicitly that the shape does not change during propagation, the intended comparison is that the profile has shifted by $2\,$m to the right in $1\,$s.

Indeed, the initial profile centered near $x=-1$ appears at $t=1$ centered near $x=1$, i.e. shifted by

$\Delta x = 2\,\text{m}$

in

$\Delta t = 1\,\text{s}$

Therefore,

$v=\frac{\Delta x}{\Delta t}=\frac{2}{1}=2\,\text{m/s}$

4. Final answer

$\boxed{2}$