jmc

We let the side length of the base of the box be $x$, so the height of the box is $x+3$. The surface area of each box then is $2x^2+4(x)(x+3)$. Therefore, we must have $$2x^2+4x(x+3)\ge 90.$$Expanding the product on the left and rearranging gives $2x^2+4x^2+12x-90\ge 0$, so $6x^2+12x-90\ge 0$. Dividing by 6 gives $x^2+2x-15\ge 0$, so $(x+5)(x-3)\ge 0$. Therefore, we have $x\le -5\text{ or }x\ge 3$. Since the dimensions of the box can't be negative, the least possible length of the side of the base is $3$. This makes the height of the box $3+3=\boxed{6}$.

Note that we also could have solved this problem with trial-and-error. The surface area increases as the side length of the base of the box increases. If we let this side length be 1, then the surface area is $2\cdot 1^2+4(1)(4)=18$. If we let it be 2, then the surface area is $2\cdot 2^2+4(2)(5)=8+40=48$. If we let it be 3, then the surface area is $2\cdot 3^2+4(3)(6)=18+72=90$. So, the smallest the base side length can be is 3, which means the smallest the height can be is $\boxed{6}$.