According to the question it is given that,

Depth of the cylindrical vessel = Height of the cylindrical vessel $=h=42cm$ (common for both)

Inner diameter of the cylindrical vessel $=14cm$

Thus, the inner radius of the cylindrical vessel $={{r}_{1}}=14/2=7cm$

Outer diameter of the cylindrical vessel $=16cm$

Now, the outer radius of the cylindrical vessel $={{r}_{2}}=16/2=8cm$

Then,

The volume of the cylindrical vessel

$V=\pi \left( r_{2}^{2}-r_{1}^{2} \right)\times h$

$=\pi \left( {{8}^{2}}-{{7}^{2}} \right)\times 42$

$=22/7\times 15\times 42$

$V=1980c{{m}^{3}}$

Hence, the total space between the two vessels is $1980c{{m}^{3}}$, which is also the amount of cork dust required.