Solution:

Here,
the length of the block (l) = 18 cm
breadth of the block (b) = 12 cm 
thickness of the block (h) = 8 cm
Now,
the surface area of the block = 2 (l b + bh + l h)
= 2 (18 × 12 + 12 × 8 + 18 × 8) cm² 
= 2 (216 + 96 + 144) cm²     
= 912 cm² 
Again,
Volume of the block = l × b × h 
= 18 cm × 12 cm × 8 cm
= 1728 cm³ 
∴ The volume of the block is 1728 cm³

Solution:

Here,
Volume of the box (V) = 1600 cm³ 
height of the box = 5 cm
Now,
Volume of the box = Area of its base × height 
∴ Area of its base × 5 = 1600
or, Area of its base × height = 1600 
or, Area of its base = \(\frac{1600}{5}\) 
or, Area of its base = 320 cm²
So, its base covers an area of 320 cm² on the table.

Solution:

Here,
the surface area of the cubic block = 96 cm²
or, 6 l2 = 96 cm²
or, l² = \(\frac{96}{6}\) cm² 
or, l² = 16 cm² 
or, l = \(\sqrt{16 cm^2}\)
or, l = 4 cm
∴ The length of its each edge is 4 cm.

Solution:

Here,
Volume of the cubical room = Volume of air 
∴ Volume of the cubical room = 512 m³ 
or, l³ = 512 m³ 
or, l = \(\sqrt[3]{512 m^3}\)
or, l = 8 m
So, the length of its floor is 8 m.

Solution:

Let, the breadth of the cuboid be x cm.
∴ The length of the cuboid will be 2x cm.
Now,
Volume of the cuboid = 768 cm³ 
or, l × b × h = 768 cm³ 
or, 2x × x × 6 cm = 768 cm³ 
or, 2x² = \(\frac{768 cm^3}{2}\) 
or, 2x² = 128 cm² 
or, x² = \(\frac{128}{2}\) cm² 
or, x² = 64 cm²
or, x = \(\sqrt{64 cm^2}\) 
or, x = 8 cm
So, the breadth (b) = x = 8 cm and the length (l) = 2x 
= 2 × 8 cm
= 16 cm
Again, 
The surface area of the cuboid = 2 (l b + bh + l h)
= 2 (16×8 + 8×6 + 16×6) cm²
= 2 (128 + 48 + 96) cm² 
= 2 × 272 cm² 
= 544 cm² 
∴ The surface area of the cuboid is 544 cm²

Solution:

length of the block (l) = 6 cm 
breadth of the block (b) = 8 cm
 thickness of the block (h) = 4 cm
Now,
Volume of the block = l × b × h 
= 16 cm × 8 cm × 4 cm 
= 512 cm³ 
volume of the cube = volume of the block 
or,  l³ = 512 cm³ 
or, l = \(\sqrt[3]{512 cm^3}\) 
or, l = 8 cm
Again, 
the surface area of the cube = 6 l² 
= 6 × (8 cm)² 
= 384 cm² 

Solution:

Here,
length of the rectangular box (l) = 24  cm
breadth of the rectangular box (b) = 12 cm
height of the rectangular box (h) = 16 cm
∴ Volume of the rectangular box = l × b × h 
= ( 24 × 12 × 16) cm³
= 4608   cm³ 
Also,
volume of milk powder = volume of the box = 4608 cm³ 
Again,
volume of each cubical box = (8 cm)³ = 512 cm³ 
Now, 
the required number of cubical boxes = \(\frac{volume \; of \; milk \; powder}{volume \; of \; each \; cubical \; box}\) 
= \(\frac{4608}{512}\) 
= 9
So, 9 cubical boxes are required to empty the milk powder from the box.

Solution:

Here,
the volume of water = 3000 l
= 3000 × 1000 cm³
Now,
the volume of the part of the tank containing water = volume of water 
or, Area of its base × height = 3000 × 1000 cm³ 
or, 30,000 cm² × h = 3000 × 1000 cm³
or, h = \(\frac{3000 × 1000 cm^3}{30,000 cm^2}\)
or, h = 100 cm
So, the required height of water level in the tank is 100 cm (or 1 m).