Solution: the conventional approach

Assume that during the week, he would have made w blocks. [As Tommy arranges these blocks in a cube, w = c^3.]

In the next two days, he would have made w/3 blocks. [As Tommy arranges these w + w/3 = 4*w/3 blocks in three squares, (4*w/3)/3 = 4*w/9 is a perfect square = s^2.]

Now, 4*w/9 = s^2 means w = (9*s^2)/4. Now, w = c^3. Equating the two, (s^2)/(c^3) = 4/9.

Now comes the most difficult part. We have to find the least among the sets of values for s and c for which 9*s^2 = 4*c^3 would be true.

The least set of such values for s and c would be 18 and 9, as 9*18^2 = 2916 = 4*9^3. (There are other sets like 144 and 36 etc, but they are larger.)

Thus w = c^3 = 729. Also, since w + w/3 = 4*w/3 constitutes half the order, the order must be for twice that number, namely, 8*w/3 = 8*729/3 = 1944.

You may like to cross-check with the Smart Scholars Method

1