Four identical cones each of radius 10.5 cm and height 14 cm are cut from a cuboid of dimensions $$30 cm \times 32 cm \times 40 cm$$ (base of each cone lies on the surface of cuboid). What is the total surface area (in cm$$^2$$) of the remaining solid?

The surface $$S_{cone}$$ of a cone can be divided into two parts, the slanting surface $$S_{slant}$$ and the base disc surface $$S_{disc}$$.

$$S_{cone}=S_{slant}+S_{cone}.$$

When you cut out a cone from a cuboid, assuming you cut it out such that the base disc of the cone coincides with one of the surfaces of the cuboid, the surface of the cuboid loses the area coinciding with the base disc, but gains the slanting area. If you do this four times, the final surface area of the remaining solid is

$$S_{cuboid}+4S_{slant}-4S_{disc}.$$

So, $$4S_{slant}=4\pi r\sqrt{r^2+h^2}=4\times\frac{22}{7}\times10.5\sqrt{10.5^2+14^2}=2310.$$

And, $$4S_{disc}=4\pi r^2=4\times\frac{22}{7}\times10.5^2=1386.$$

And, $$S_{coboid}=2\left(lb+lh+bh\right)=2\left(30\times40+30\times32+32\times40\right)=6880.$$

So, required surface area = $$6880+2310-1386=7804\ cm^2.$$

B is correct choice.