A well of diameter 3 m is dug 14 m deep

The inner radius of the well is $\frac{3}{2} m$ and the height is $14 m$. Therefore, the volume of the Earth taken out of it is

$V_{1}=\pi \times\left(\frac{3}{2}\right)^{2} \times 14 \mathrm{~m}^{3}$

The inner and outer radii of the embankment are $\frac{3}{2} m$ and $4+\frac{3}{2}=\frac{11}{2} m$ respectively. Let the height of the embankment be $h$. Therefore, the volume of the embankment is

$V_{2}=\pi \times\left\{\left(\frac{11}{2}\right)^{2}-\left(\frac{3}{2}\right)^{2}\right\} \times h \mathrm{~m}^{3}$

Since, the volume of the well is same as the volume of the embankment; we have

$V_{1}=V_{2}$

$\Rightarrow \pi \times\left(\frac{3}{2}\right)^{2} \times 14=\pi \times\left\{\left(\frac{11}{2}\right)^{2}-\left(\frac{3}{2}\right)^{2}\right\} \times h$

$\Rightarrow h=\frac{9 \times 14}{112}$

$\Rightarrow h=\frac{9}{8} \mathrm{~m}$

Hence, the height of the embankment is $h=\frac{9}{8} \mathrm{~m}$