After 3 minutes the volume is $\displaystyle 390 - 3 \cdot 60~cm^3 = 210~cm^3$.
The volume of a cone is $\displaystyle V = \frac{\pi}{3}r^2 h$.
The 45 degree angle gives $\displaystyle r = h~\Rightarrow~V = \frac{\pi}{3}h^3~\Leftrightarrow~h = \left(\frac{3 V}{\pi}\right)^{1/3}$.
$\displaystyle \frac{dh}{dt} = \frac{dh}{dV}\cdot\frac{dV}{dt} = \frac{1}{\left(9\pi V^2\right)^{1/3}}\frac{dV}{dt} \approx 0.028~cm/s$.
Yeah, that's a bit strange. Try showing her the expression
$\displaystyle \frac{dh}{dt} = \frac{1}{\left(9\pi\left(V_0 + t\cdot \frac{dV}{dt}\right)^2\right)^{1/3}}\cdot\frac{dV}{dt}$
where $\displaystyle V_0$ is the initial volume in $\displaystyle cm^3$, $\displaystyle t$ is the time in seconds and $\displaystyle \frac{dV}{dt}$ is the change in volume w.r.t. time $\displaystyle [cm^3/s]$, and ask her if she gets the same. Because if it's true, then you get the correct answer when inserting 3 mins and $\displaystyle 2~cm^3/s$.