first off, let's notice something, the graph seems a bit misleading, since the height of the nut is 0.5 cm, whilst the side of the base is 0.6 cm, however in the picture 0.5 appears longer. That said
1)
the volume of the nut itself is simply the volume of a hexagonal prism, which will just be the product of the area of the hexagon and the height.
![\bf \textit{area of a regular polygon}\\\\ A=\cfrac{1}{2}ap~~ \begin{cases} a=apothem\\ p=perimeter\\ \cline{1-1} a=0.5\\ p=\stackrel{0.6\times 6}{3.6} \end{cases}\implies A=\cfrac{(0.5)(3.6)}{2}\implies A=0.9 \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{volume of the hexagonal prism}}{\stackrel{\textit{hexagon's area}}{(0.9)}~~\stackrel{\textit{height}}{(0.5)}\implies 0.45}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Barea%20of%20a%20regular%20polygon%7D%5C%5C%5C%5C%20A%3D%5Ccfrac%7B1%7D%7B2%7Dap~~%20%5Cbegin%7Bcases%7D%20a%3Dapothem%5C%5C%20p%3Dperimeter%5C%5C%20%5Ccline%7B1-1%7D%20a%3D0.5%5C%5C%20p%3D%5Cstackrel%7B0.6%5Ctimes%206%7D%7B3.6%7D%20%5Cend%7Bcases%7D%5Cimplies%20A%3D%5Ccfrac%7B%280.5%29%283.6%29%7D%7B2%7D%5Cimplies%20A%3D0.9%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bvolume%20of%20the%20hexagonal%20prism%7D%7D%7B%5Cstackrel%7B%5Ctextit%7Bhexagon%27s%20area%7D%7D%7B%280.9%29%7D~~%5Cstackrel%7B%5Ctextit%7Bheight%7D%7D%7B%280.5%29%7D%5Cimplies%200.45%7D)
2)

3)
well, the composite figure is just a hexagonal nut with a cylindrical hole, so if we simply get the volume of the prism and subtract the volume of the cylindrical hole, what's leftover, is the volume of the nut alone without the hole.

4)
in short, dividing the mass of 3.03 by our result from 3)
3.03 ÷ 0.2 = 1.515.