Determine the speed at which the volume of an object shrinks to...

Let the rest volume of the object be $V_0 = L_x L_y L_z$.

When the object moves with a speed $v$ along the x-axis, its length in the direction of motion contracts due to length contraction, while the perpendicular dimensions remain unchanged.

The new length is $L'_x = L_x \sqrt{1 - \dfrac{v^2}{c^2}}$.

The new volume is $V = L'_x L_y L_z = V_0 \sqrt{1 - \dfrac{v^2}{c^2}}$.

Given that the volume shrinks to half of its rest value, $V = \dfrac{V_0}{2}$.

$\dfrac{V_0}{2} = V_0 \sqrt{1 - \dfrac{v^2}{c^2}}$

$\dfrac{1}{4} = 1 - \dfrac{v^2}{c^2}$

$\dfrac{v^2}{c^2} = \dfrac{3}{4}$

$v = \dfrac{\sqrt{3}\,c}{2}$