smc

Let $x,y$, and $z$ be the unique lengths of the edges of the box. Each box has $4$ edges of each length, so: $$4x+4y+4z=140⟹x+y+z=35.$$ The spacial diagonal (longest distance) is given by $\sqrt{x^2+y^2+z^2}$. Thus, we have $\sqrt{x^2+y^2+z^2}=21$, so $x^2+y^2+z^2=21^2$. Our target expression is the surface area of the box: $$S=2xy+2xz+2yz.$$ Since $S$ is a symmetric polynomial of degree $2$, we try squaring the first equation to get: $$35^2=(x+y+z{)}^2=x^2+y^2+z^2+2xy+2yz+2xz=35^2.$$ Substituting in our long diagonal and surface area expressions, we get: $21^2+S=35^2$, so $S=(35+21)(35-21)=56\cdot 14=784$, which is option $\boxed{784}$.