A well of diameter 3M is …

The diameter of well = 3 m
{tex}\therefore {/tex} The radius of well {tex}(r) = \frac{3}{2}{/tex}m and Depth of earth dug (h) = 14 m
Width of the embankment = 4 m
{tex}\therefore {/tex} The radius of the well with embankment r' = {tex}\frac{3}{2} + 4 = \frac{{11}}{2}{/tex}m
Let the height of the embankment be h'm
According to the question,
The volume of embankment = Volume of the earth dug
{tex} \Rightarrow \pi \left[ {{{(r')}^2} - {r^2}} \right]h' = \pi {r^2}h{/tex}
{tex} \Rightarrow \left[ {{{\left( {\frac{{11}}{2}} \right)}^2} - {{\left( {\frac{3}{2}} \right)}^2}} \right]h' = {\left( {\frac{3}{2}} \right)^2} \times 14{/tex}

 {tex} \Rightarrow \left[ {\frac{{121}}{4} - \frac{9}{4}} \right]h' = \frac{9}{4} \times 14{/tex}
{tex} \Rightarrow \frac{{112}}{4} \times h' = \frac{9}{4} \times 14{/tex}

 {tex} \Rightarrow h' = \frac{{9 \times 14 \times 4}}{{4 \times 112}}{/tex}
{tex} \Rightarrow h' = 1.125m{/tex}