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 $9/8m$