If the length of one side of the base is s. Then the height is 3s.
V=area of base*height=s^2*(3s)=3s^3
So at time t, V(t)=3*s(t)^3, where V(t) and s(t) are volume and length of the base at time t.
Now differentiate with respect to t:
dV/dt=9s^2*(ds/dt) {by the chain rule}
2=9*(8^2)*ds/dt
ds/dt=2/(9*64)=1/288 cm/s