#### Answer

$6.3\dfrac {m}{s^{2}}$

#### Work Step by Step

Lets assume that $H_{0}$ is initial height of the box relative to the ground.
Total energy of the spring and box system initially will be :
$E_{tot}=mgH_{0}..........(1)$
When the box starts to move, its potential energy will be converted to the box’s kinetic energy and spring’s potential energy. Lets assume that box will descend distance $L$ on the incline (so the spring will stretch by distance $L$). Now, lets write total energy equation again:
$E_{tot}=E_{spring}+E_{box}=\dfrac {kL^{2}}{2}+\dfrac {mv^{2}_{1}}{2}+mgH_{1}.........(2)$
Here, $H_{1}$ is the height of the box relative to the ground after descending $L$ distance. From the incline, we get:
$\Delta H=H_{0}-H_{1}=L\sin \theta \left.......( 3\right) $
From (1) and (2), we get:
$mg\left( H_{0}-H_{1}\right) -\dfrac {kL^{2}}{2}=\dfrac {mv^{2}_{1}}{2}\Rightarrow v_{1}=\sqrt {2g\left( Ho-H_{1}\right) -\dfrac {k}{m}L^{2}}\left...( 4\right) $
From (3) and (4) we get:
$v_{1}=\sqrt {L\left( 2g\sin \theta -\dfrac {kL}{m}\right) }....(5)$
So if the block momentarily stops, then $v_{1}=0\left.....( 6\right) $
From (5) and (6), we get:
$L=\dfrac {2mg\sin \theta }{k}\approx 0.21m...........(7)$
So at that point, lets find the acceleration of the box:
$ ma=kL-mg\sin \theta $
Using (7), we get $a=g\sin \theta \approx 6.3\dfrac {m}{s^{2}} $