This is the context of the question: 
I'm assuming by the space diagonal (although I'm not sure) to be the area of the right-angled traingle created by the diagonal. Let this space be $V$, then we will have $V=\frac{1}{2}\sqrt{2}x^2$, where $x$ is the lenght of the edges at time $t$. Now, $\frac{dV}{dx}=\sqrt{2}x$, and $\frac{dV}{dt}=\frac{dV}{dx}\frac{dx}{dt}=2\sqrt{2}x$. But, we haven't directly been told $x$ abd I can't think of anyway to find it. Can anyone suggest what I do?
(The answer is supposed to be $-2\sqrt{3} cm/s$)
Your assumption is wrong. The space diagonal is the line connecting two vertices of the cube not belonging to the same face. So it is the hypothenuse of a rectangle triangle whose edges are $a$ and $a\sqrt{2}$ (diagonal of one of the faces), where $a$ is the edge length. This means that $sd=a\sqrt{3}$ and therefore the rate of decrease is $\dot{a}\sqrt{3}=-2\sqrt{3}$ as expected.