Step 1: Get the sample volume and the water mass lost on drying.
$V = \dfrac{\pi}{4}(0.06)^2(0.1) \approx 2.827\times10^{-4}$ m$^3$, and the moist sample lost $0.48-0.44=0.04$ kg of water on drying.
Step 2: Work directly in volumetric water content instead of depths.
Volume of that water: $V_w = 0.04/1000 = 4\times10^{-5}$ m$^3$.
Volumetric water content is just this water volume as a fraction of the total sample volume:
\[ \theta = \frac{V_w}{V} = \frac{4\times10^{-5}}{2.827\times10^{-4}} \approx 0.1415 \]
Step 3: Use the fact that volumetric water content equals depth of water per unit depth of soil.
For any soil column of uniform cross-section, $\theta$ already carries the units of (m water)/(m soil), since both the water volume and the soil volume share the same cross-sectional area. So no separate depth calculation is needed.
Final Answer:
The volumetric water content, and hence the water depth per metre of soil, comes out to about $0.14$.
\[ \boxed{\approx 0.14} \]