Answer
$a = -200~m/s^2$
Work Step by Step
In part (b), we found that the mass of the block is $0.20~m$
At $t = 0.10~s$, the velocity is $v = 0$ so the block must be at one of the endpoints.
Just after this time the velocity is then negative, so the block must be at the most positive displacement at $t = 0.10~s$
We can find the acceleration:
$F = ma$
$-kx = ma$
$a = \frac{-kx}{m}$
$a = \frac{-(200~N/m)(0.20~m)}{0.20~kg}$
$a = -200~m/s^2$
