As per the question it is given that,
Diameter of the well $=3m$
Then, the radius of the well $=3/2m=1.5m$
Depth of the well (h) $=14m$
Width of the embankment (thickness) $=4m$
Therefore, the radius of the outer surface of the embankment $=(4+1.5) m=5.5m$
Assume the height of the embankment be taken as h m
As we know that the embankment is a hollow cylinder
Volume of the embankment $=\pi \left( {{R}^{2}}-{{r}^{2}} \right)\times h$
$=\pi \left( {{5.5}^{2}}-{{1.5}^{2}} \right)\times h$ ….. (i)
Volume of earth dug out
$=\pi \times {{2}^{2}}\times 14$ ….. (ii)
On equating both (i) and (ii) we get,
$\pi \left( {{5.5}^{2}}-{{1.5}^{2}} \right)\times h=\pi \times {{\left( 3/2 \right)}^{2}}\times 14$
$\left( 30.25-2.25 \right)\times h=9\times 14/4$
$h=9\times 14/\left( 4\times 28 \right)$
$h=9/8m$
Hence, the height of the embankment is $9/8m$