Answer
Decrease by the rate $\frac{2}{27} \approx 0.07 \text{cm}/\text{min}$
Work Step by Step
The volume of a cube with edge $a$ is:
$$V=a^{3}$$
Using the chaine rule it follows:
$$\frac{dV}{dt}=\frac{dV}{da}\cdot \frac{da}{dt}$$
So the derivative of the volume is:
$$\frac{dV}{da}=3a^{2}$$
so:
$$\frac{dV}{dt}=3a^{2}\cdot \frac{da}{dt}$$
Since the volume of the cube decreases it follows that $\frac{dV}{dt} \lt 0$ so:
$$-2=3\cdot (3)^{2}\cdot \frac{da}{dt}$$
$$\frac{da}{dt}=-\frac{2}{27} \approx -0.07 \text{cm}/\text{min}$$
