Let the side length of the kth cube be A(k), then A( 1)=2, and a (k+1) = a (k) x √ 2/2; Therefore A(n)=2x.

(√2/2)^(n- 1)；

Assuming that the number of required cubes is n, it is obvious that when the tower is stacked on the second floor, the total surface area is twice that of two cubes minus the area of one side of the second floor cube, and when stacked on the third floor, the total surface area is twice that of three cubes minus the area of one side of the second floor cube, minus the area of one side of the third floor cube, and so on:

When stacked to the nth layer, the surface area of the tower is: the surface area of each layer of cubes minus the 2nd, 3rd,... nth cubes twice, that is:

6[A( 1)xA( 1)+A(2)xA(2)+……A(n)xA(n)]-2[A(2)xA(2)+……A(n)xA(n)]

= 4[A( 1)xA( 1)+A(2)xA(2)+……A(n)xA(n)]+2[A( 1)xA( 1)]

=4[4+4x( 1/2)+4x( 1/2)^2……+4x( 1/2)^(n- 1)]+2x2x2

=4x4x[ 1+( 1/2)+( 1/2)^2+……+( 1/2)^(n- 1)]+8

= 16[2-( 1/2)^n]+8>； 39

Solve the above inequality to get n>5;

So the value of n should be at least 6.