A rectangular field is 70 m long and 60 m broad.

Dimension of the well $=14 \mathrm{~m} \times 8 \mathrm{~m} \times 6 \mathrm{~m}$

The volume of the dug $-$ out earth $=14 \times 8 \times 6=672 \mathrm{~m}^{3}$

Now, we will spread this dug $-$ out earth on a field whose length, breadth and height are $70 \mathrm{~m}, 60 \mathrm{~m}$ and $h \mathrm{~m}$, respectively.

Volume of the dug $-$ out earth $=$ length $\times$ breadth $\times$ height $=70 \times 60 \times h$

$\Rightarrow 672=4200 \times h$

$\Rightarrow h=\frac{672}{4200}=0.16 \mathrm{~m}$

We know that $1 \mathrm{~m}=100 \mathrm{~cm}$

$\therefore$ The earth level will rise by $0.16 \mathrm{~m}=0.16 \times 100 \mathrm{~cm}=16 \mathrm{~cm} .$