20) D

It is advisable to skip such long and tricky questions and finish the other simpler ones first.

We need to use a lot of Pythagoras in this question.

Side of smaller cube = 1cm
Side of 2nd cube = A
Side of 3rd cube = B

5 faces of the smallest cube can be seen.
Surface area of the smallest cube = 1 x 1 x 5 = 5 cm²

The vertices of the smallest cube are at the midpoints of A.
You can see that multiple right-angled triangles are formed.

1² = (½A)² + (½A)²
1 = ¼A² + ¼a²
1 = ½a²
2 = A²
A = √2

4 faces of the 2nd cube can be seen and the top face is partially covered by the smallest cube.
Surface area = 5 x √2 x √2 = 10cm²
Area covered by one of the faces of the smaller cube = 1 x 1 = 1cm²
So total surface area of the 2nd cube = 10 – 1 = 9cm²

Again the vertices of the second cube are at the midpoints of B.
You can see that multiple right-angled triangles are formed.

(√2)² = (½B)² + (½B)²
2 = ¼B² + ¼B²
2 = ½B²
4 = B²
B = √4 = 2cm

5 faces of the largest cube are uncovered and 1 is partially covered.
Surface area = 6 x 2 x 2 = 24cm²
Again the top face is partially covered by the 2nd cube.
Surface area of one of the faces of the 2nd cube = √2 x √2 = 2cm²
Total surface area of largest cube = 24 – 2 = 22cm²

Total surface area of solid = 22 + 9 + 5 = 36 cm²