#### Answer

direction of acceleration of the box is upwards

#### Work Step by Step

Lets assume that $H_{0}$ is initial height of the box relative to the ground. The 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 the spring’s potential energy. Lets assume that the box will descend a distance $L$ on the incline (so the spring will stretch by a distance $L$) and lets write total energy equation again:
$E_{tot}=E_{spring}+E_{box}=\dfrac {kL^{2}}{2}+\dfrac {mv^{2}_{1}}{2}+mgH_{1}...(2)$
$H_{1}$ is the height of the box relative to the ground after descending a distance $L$. 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 block momentarily stops then $v_{1}=0\left.....( 6\right) $
From (5) and (6), we get $L=\dfrac {2mg\sin \theta }{k}\approx 0,21m.....(7)$
Upward force acting on the box on incline is $F_{up}=k\times L=2mg\sin \theta $ And the downward foce on box is: $F_{down}=mg\sin \theta $. So we can see $F_{up} > F_{down}$ so the direction of acceleration of the box is upwards on the incline.