A well with inner radius 4 m is dug 14 m deep.

The inner radius of the well is 4m and the height is 14m. Therefore, the volume of the Earth taken out of it is

$V_{1}=\pi \times(4)^{2} \times 14 \mathrm{~m}^{3}$

The inner and outer radii of the embankment are 4m and 4+3=7m respectively. Let the height of the embankment be h. Therefore, the volume of the embankment is

$V_{2}=\pi \times\left\{(7)^{2}-(4)^{2}\right\} \times h \mathrm{~m}^{3}$

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

$V_{1}=V_{2}$

$\Rightarrow \pi \times(4)^{2} \times 14=\pi \times\left\{(7)^{2}-(4)^{2}\right\} \times h$

$\Rightarrow h=\frac{(4)^{2} \times 14}{33}$

$\Rightarrow h=6.78 \mathrm{~m}$

Hence, the height of the embankment is $6.78 \mathrm{~m}$