Four cubic blocks with edge $ 4 $ cm were kept two on top of two and fused together into a block. Find the total surface area of the block and the cost of painting it at $ Rs.32 $ per $ c{m^2} $
A) $ Rs.8992 $
B) $ Rs.8092 $
C) $ Rs.8192 $
D) $ Rs.8190 $

Hint: The dimensions length, breadth and height of a single cube are $ 4 $ cm each. After they are mounted, they would form a cuboid so calculate the dimensions of this new formed cuboid. The total surface area of a cuboid is equal to $ 2\left( {lb + bh + lh} \right) $

Complete step-by-step answer:
Given to us are four cubes with dimensions equal to $ 4 $ cm.
 When these four cubes are mounted such that two cubes are on top of another two, they form a cuboid. So now we find the dimensions of this new cuboid.
Length- Length of this cuboid will be the sum of lengths of two cubes i.e. $ 4 + 4 $ cm.
Hence, the length is $ l = 8 $ cm.
Breadth - Since the cubes are mounted vertically, the breadth remains same and hence $ b = 4 $ cm.
Height- Since two cubes are mounted on top of another two, the height would be the sum of heights of two cubes. Hence the height is $ h = 4 + 4 = 8 $ cm.
Now, we have to calculate the total surface area of this cuboid. The formula for total surface area is $ T.S.A = 2\left( {lb + bh + lh} \right) $
By substituting the calculated values of the dimension in this formula, we get $ T.S.A = 2\left( {8 \times 4 + 4 \times 8 + 8 \times 8} \right) $
On solving, $ 2\left( {128} \right) = 256 $
Therefore the total surface area is $ 256c{m^2} $
Now, the cost of painting one $ c{m^2} $ is given as $ Rs.32 $
The cost of painting $ 256\;c{m^2} $ would be $ Rs.256 \times 32 = Rs.8192 $
So, the correct answer is “Option C”.

Note: It is to be noted that the units should always be mentioned when writing any dimension. In this case the units of length, breadth and height are centimeters whereas the units of total surface area are $ c{m^2} $ and the cost is mentioned in rupees.