From a solid cylinder of height 14 cm and base diameter 7 cm, two equal conical holes each of radius 2.1 cm and height 4 cm are cut off.

We have,

the height of the cylinder, $H=14 \mathrm{~cm}$,

the base radius of cylinder, $R=\frac{7}{2} \mathrm{~cm}$,

the base radius of each conical holes, $r=2.1 \mathrm{~cm}$ and

the height of each conical holes, $h=4 \mathrm{~cm}$

Volume of the remaining solid = Volume of the cylinder - Volume of 2 conical holes

$=\pi R^{2} H-2 \times \frac{1}{3} \pi r^{2} h$

$=\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 14-\frac{2}{3} \times \frac{22}{7} \times 2.1 \times 2.1 \times 4$

$=539-36.96$

$=502.04 \mathrm{~cm}^{3}$

So, the volume of the remaining solid is 502.04 cm3.