Same Height, Same Area, Different Volumes

Let $a,b$ be the width and length of the bigger box and $c,d$ be that of the other box respectively.

The total surface area = $166 = 2(ab + a+ b) = 2(cd+c+d)$ .

$83 = ab + a+ b = cd+c+d$ .

The volume difference = $1 = ab-cd$ . $ab = cd+1$

Hence, $a+b+1 = c+d$ . $1 = (c+d)-(a+b)$ .

Now let us rewrite the equation in a more composite form:

$83 = ab + a+ b = (a+1)(b+1)-1$ . $84 = (a+1)(b+1)$

$83 = cd + c+ d = (c+1)(d+1)-1$ . $84 = (c+1)(d+1)$

Thereby, we are trying to find the factors of $84$ , whose pair sums are different by $1$ , and we will obtain $84 = 7\times 12 = 6\times 14$ since $(6+14)-(7+12) = 1$ .

As a result, $a=7-1 = 6$ ; $b=12-1=11$ . $c = 6-1 = 5$ ; $d=14-1 =13$ .

Checking the answers: $2(6\cdot 11 + 6 + 11) = 166 = 2(5\cdot 13 + 5 + 13)$ .

Then the volume of the bigger box = $6\cdot 11 = \boxed{66}$ and the volume of the smaller box = $5\times 13 = 65$ .