Calculate the spring constant of oscillating mass on a spring colliding with a wall

[Aurelius120]

But a rather curious coincidence:
If I go by my first interpretation:

My interpretation was that the given diagram was of the relaxed spring (at $0$) which would be compressed to left $(-A)$ and released. The wall being on the right $(+A/2)$. That means one oscillation will be $$(-A)\rightarrow 0\rightarrow \frac{+A}{2} \rightarrow (-A)$$

Period of motion =0.2s given in question is time for $(-A)\rightarrow 0 \rightarrow \frac{+A}{2} \rightarrow (-A)$
Then $$-A\cos (\omega t)=\frac{+A}{2}$$
$$\implies -\sqrt 2 \cos (\omega t)=\frac{+1}{\sqrt 2}$$
$$\implies \cos (\omega t)=\frac{-1}{2} \implies \omega t=\frac{2\pi}{3}$$
According to question time period of oscillation, 2t=0.2s therefore $$2\omega t= 0.2\omega = \frac{4\pi}{3}$$
This is the same incorrect answer I obtained in POST(1,11,22) and therefore this method is equivalent to that.

Period of motion =0.2s given in question is time for$(-A)\rightarrow 0 \rightarrow \frac{+A}{2}$
Here I take t = 0.2s

Then, $$\omega t=0.2 \omega=\frac{2\pi}{3}\implies \frac{k}{0.9}=\frac{4\pi^2}{0.6^2}$$
Now this gives the correct answer.

This could either be a coincidence or even my initial interpretation is correct if period of motion in the question refers to time taken before first collision. [(-A) to (+A/2)]Could this question have two meanings? (POST 28)

 

[haruspex]

before first collision. [(-A) to (+A/2)]

For the third time, it is not [(-A) to (+A/2)].
"Spring compressed to √2cm. "
A = √2cm.
"Block at 1/√2cm from wall."
Block starts at A/2 from wall; equilibrium is A from starting position, so A/2 to the right of the wall. The motion is an oscillation between -A and -A/2.