3x3

If we label the 3x3 grid as shown above, then we can determine some of the values. To begin with, $abc = 1$ and $def = 1$ . By multiplying these equations, we get that $abcdef = 1$ .

Now we know that $abde = 2$ as it is a 2x2 square. $abcdef = 1$ divided by $abde = 2$ gives us $cf =$ $\frac{1}{2}$ . However, as $cfi$ must equal $1$ , this means $i = 2$ .

However, our logic can be applied to any 90 degree rotation of the grid, as the grid will still have columns, rows, and 2x2 boxes which equal the same. Therefore, all of the corner boxes (a,c,g and i) = 2.

This means that, for example, $abc = 1$ so $4b = 1$ and $b =$ $\frac{1}{4}$ . Again, the same logic can be applied for squares (b,d,f and h).

Therefore, as $def = 1$ , $e$ must equal 16. By checking, we see that all of the conditions have been satisfied and we have proved that this grid below is the only grid possible (note that rotating the grid gives the same result).

This means that the answer to the problem is $\frac{1}{4}$

$a_{11} = \frac 1 {a_{12}\cdot a_{13}} = \frac 1 {1/a_{22}a_{32}\cdot 1/a_{23}a_{33}} = a_{22}a_{23}a_{32}a_{33} = 2,$ and the same applies to all corners: $a_{11} = a_{13} = a_{31} = a_{33} = 2$ . Then $a_{12} = \frac 1 {a_{11}\cdot a_{13}} = \frac 1 {2^2} = \frac 14,$ and the same applies to all edges: $a_{12} = a_{32} = a_{21} = a_{23} = 1/4$ . Finally, $a_{22} = \frac 1{a_{12}\cdot a_{32}} = \frac 1 {(1/4)^2} = 16.$ Thus we have determined the entire square; it only contains the numbers $2, \boxed{1/4}, 16$ .