Your answer is correct until the arithmetical error in the last line. $k=10\pi^2 N/m$, not $1000\pi^2$.

The mass performs SHM which is 'frustrated' by the wall. It does not perform complete oscillations. We can assume that the collision is "instantaneous" - which means that the collision time is much less than the period $T=0.2s$.

The amplitude of free oscillations (if the wall were not there) is $\sqrt2$, the maximum compression of the spring. The mass rebounds from the wall before it reaches the equilibrium position, since $\frac{1}{\sqrt2} \lt \sqrt2$.

The angular frequency of free oscillations is $\omega=\sqrt{\frac{k}{m}}$. Until it collides with the wall the equation of motion is $x=\sqrt2 \cos(\omega t)$ where $x$ is measured from the equilibrium position. The time $t \approx \frac12T=0.1s$ after maximum compression until the mass collides with the wall is given by

$\sqrt2-\frac{1}{\sqrt2}=\sqrt2 \cos(0.1\omega)$

$2-1=1=2\cos(0.1\omega)$

$\cos(0.1\omega)=\frac12=\cos(\frac13\pi)$

$\omega=\frac{10}{3}\pi$

$\omega^2=\frac{k}{m}=\frac{100}{9}\pi^2$

$k=\frac{100}{9}\pi^2 m = 10\pi^2 \approx 100N/m$.