Here’s what I got.
The idea here is that you need to determine the volume of the wood needed to make the box. To do that, you will have to consider two boxes
The volume of the wood used to make the box will be the diference between these two volumes.
##V_”wood” = V_”outside” – V_”inside”##
I assume that “ecteenal” is actualy external. The volume of the outside box will be
##V_”outside” = “20 cm” xx “15 cm” xx “10 cm” = “3000 cm”^3##
Now, since the wood is ##”1 cm”## thick, it follows that the inside of the box will have the following dimensions – keep in mind that you need to take thickness into account on both sides of the box!
##V_”inside” = [20-(1+1)]”cm” xx [15 – (1+1)]”cm” xx [10-(1+1])”cm”##
##V_”inside” = “18 cm” xx “13 cm” xx “8 cm” = “1872 cm”^3##
The volume of the wood will thus be
##V_”wood” = “3000 cm”^3 – “1872 cm”^3 = “1128 cm”^3##
Convert this value to cubic decimeters to make it match the units you have for the of wood
##1128color(red)(cancel(color(black)(“cm”^3))) * “1 dm”^3/(10^3color(red)(cancel(color(black)(“cm”^3)))) = “1.128 dm”^3##
Now use the given to find the mass of this volume of wood
##1.128color(red)(cancel(color(black)(“dm”^3))) * “1 kg”/(1color(red)(cancel(color(black)(“dm”^3)))) = “1.128 kg”##
I’ll leave the answer rounded to two
##m_”box” = color(green)(“1.1 kg”)##
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