Answer
Work= 0.004312 Joules
Work Step by Step
As we know $\Delta U=-W_{G}$, where $W_{g}$ is the external force who does the work on the ice flake. Then, $mg(y_{f}-y_{i})=-W_{g}$- Solving for $W_{g}$ we get
$$W_{g}=mg(y_{i}-y_{f})$$
We take $y=0$ at the edge of the flake and $y_{bottom}=-r$
The mass of the ice flake is $m=2g=0.002Kg$. The radius of the bowl is $r=h=22cm=0.22m$. (We use the radius as the height because, as the gravity is a conservative force, we only care about the initial and the final point).
Then
$W_{g}=mg(y_{i}-y_{f})=mg(0-(-r))=mgr=(0.002Kg)(9.81m/s^2)(0.22m)=0.004312J$
