When placed deep in water, an empty plastic box of mass m...

Let the buoyant force on the completely submerged box be $B$.

When the empty box of mass $m$ accelerates upward at $\dfrac{g}{6}$, the equation of motion is:

$B - mg = m$\left(\dfrac{g}{6}\right)

$B = mg +$\dfrac{mg}{6} = \dfrac{7mg}{6}

Let $M$ be the mass of the sand added. The total mass becomes $m + M$. Since the volume of the box remains the same, the buoyant force $B$ remains unchanged.

For the box to accelerate downward at \dfrac{g}{6}, the equation of motion is:

$(m + M)g - B = (m + M) \left(\dfrac{g}{6}\right)$

Substituting the value of $B$ into the equation:

$(m + M)g - \dfrac{7mg}{6} = (m + M)$\left(\dfrac{g}{6}\right)

(m + M)\left(g -\dfrac{g}{6}\right) = \dfrac{7mg}{6}

$(m + M)\left(\dfrac{5g}{6}\right) = \dfrac{7mg}{6}<br><br>$5(m + M) = 7m$<br><br>$5m + 5M = 7m$<br><br>$5M = 2m$<br><br>$M = $\dfrac{2m}{5}