Answer
$3$
Work Step by Step
1) Find Curl of $\vec{F}$
$$
\text{curl}\vec{F}=\vec{\nabla}\times\vec{F}
$$
$$
=\begin{vmatrix}
\hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}}\\[1pt]
\frac{\partial}{\partial{x}} & \frac{\partial}{\partial{y}} & \frac{\partial}{\partial{z}}\\[2pt]
z^2 & 2xy & 4y^2
\end{vmatrix}=
\begin{vmatrix}
\frac{\partial}{\partial{y}} & \frac{\partial}{\partial{z}}\\[2pt]
2xy & 4y^2
\end{vmatrix}
\hat{\text{i}}-
\begin{vmatrix}
\frac{\partial}{\partial{x}} & \frac{\partial}{\partial{z}}\\[2pt]
z^2 & 4y^2
\end{vmatrix}
\hat{\text{j}}+
\begin{vmatrix}
\frac{\partial}{\partial{x}} & \frac{\partial}{\partial{y}}\\[2pt]
z^2 & 2xy
\end{vmatrix}
\hat{\text{k}}
$$
$$
=(8y-0)\hat{\text{i}}-(0-2z)\hat{\text{j}}+(2y-0)\hat{\text{k}}
=8y\hat{\text{i}}-2z\hat{\text{j}}+2y\hat{\text{k}}
$$
$$
\therefore \text{curl}\vec{F}=8y\hat{\text{i}}+2z\hat{\text{j}}+2y\hat{\text{k}}
$$
2) Define and Parametrize some Surface $S$ with Boundary of $C$
We are free to choose any surface with its boundaries defined by the given path $C$, the positively oriented closed curve from $O$ to $P, Q, R\ \text{and back to}\ O$. We will define a plane $S$ such that $S$ passes through all of the given points. The vectors
$$\vec{OP}=\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}}\ \text{and}\ \vec{PQ}=0\hat{\text{i}} + 2\hat{\text{j}}+\hat{\text{k}}$$
both lie in $S$, so
$$
\vec{OP}\times\vec{PQ}=\vec{n}.
$$
Computing $\vec{OP}\times\vec{PQ}$,
$$
\vec{OP}\times\vec{PQ}=
\begin{vmatrix}
\hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}}\\[1pt]
1 & 0 & 0\\
0 & 2 & 1\\
\end{vmatrix}=
\begin{vmatrix}
0 & 0\\
2 & 1
\end{vmatrix}
\hat{\text{i}}-
\begin{vmatrix}
1 & 0\\
0 & 1
\end{vmatrix}
\hat{\text{j}}+
\begin{vmatrix}
1 & 0\\
0 & 2
\end{vmatrix}
\hat{\text{k}}
$$
$$
\therefore \vec{n}=-\hat{\text{j}}+2\hat{\text{k}}
$$
Note that the equation of a plane $S$ with normal vector $-\hat{\text{j}}+2\hat{\text{k}}$ is
$$
0=-y+2z \Rightarrow S=z=\frac{1}{2}y
$$
Now that $S$ is represented in terms of two variables $x$ and $y$, we can find a parametric representation and a vector representation of $S$, $\vec{r}$(x,y).
$$
S=\Big\{(x,y,z)\ |\ x=x,\ y=y,\ z=\frac{1}{2}y\ |\ 0 \leq x \leq 1,\ 0 \leq y \leq 2 \Big\}
$$
$$
\vec{r}(x,y)=x\hat{\text{i}}+y\hat{\text{j}}+\frac{1}{2}y\hat{\text{k}}
$$
3) Use Stokes' Theorem to Find Work Done
Recall that
$$
\text{work}=\oint_C \vec{F}\cdot d\vec{r}
$$
Therefore, from Stokes' Theorem,
$$
\text{work}=\iint \limits_S (\text{curl}\vec{F}(x,y,z)) \cdot\vec{n} dS= \iint \limits_S
(\text{curl}\vec{F} \cdot \vec{n}) dS=
$$
$$
\iint \limits_R (\text{curl} \vec{F}(x,y) \cdot (\vec{r}_x \times \vec{r}_y)) dydx
$$
(where $R$ is the projection of $S$ in the $xy$-plane)
Parametrize $\text{curl}\vec{F}$ according to the parametrization of $S$ to find $\text{curl}\vec{F}(x,y)$
$$
\text{curl}\vec{F}(x,y)=8y\hat{\text{i}}+y\hat{\text{j}}+2y\hat{\text{k}}
$$
\noindent Compute $\text{curl} \vec{F}(x,y) \cdot (\vec{r}_x \times \vec{r}_y)$
$$
\vec{r}_x=\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}}
$$
$$
\vec{r}_y=0\hat{\text{i}}+\hat{\text{j}}+\frac{1}{2}\hat{\text{k}}
$$
$$
\vec{r}_x \times \vec{r}_y =
\begin{vmatrix}
\hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}}\\
1 & 0 & 0\\
0 & 1 & \frac{1}{2}
\end{vmatrix}=
\begin{vmatrix}
0 & 0\\
1 & \frac{1}{2}
\end{vmatrix}\hat{\text{i}}-
\begin{vmatrix}
1 & 0\\
0 & \frac{1}{2}
\end{vmatrix}\hat{\text{j}}+
\begin{vmatrix}
1 & 0\\
0 & 1
\end{vmatrix}\hat{\text{k}}
$$
$$
\therefore \vec{r}_x \times \vec{r}_y=0\hat{\text{i}}-\frac{1}{2}\hat{\text{j}}+\hat{\text{k}}
$$
$$
\text{curl} \vec{F}(x,y) \cdot (\vec{r}_x \times \vec{r}_y)=(8y\hat{\text{i}}+y\hat{\text{j}}+2y\hat{\text{k}}) \cdot (0\hat{\text{i}}-\frac{1}{2}\hat{\text{j}}+\hat{\text{k}})
$$
$$
\therefore \text{curl} \vec{F}(x,y) \cdot (\vec{r}_x \times \vec{r}_y)=0-\frac{1}{2}y+2y=\frac{3}{2}y
$$
Plug $\text{curl} \vec{F}(x,y) \cdot (\vec{r}_x \times \vec{r}_y)$ into our work equation from earlier:
$$
\text{work}=\int_0^1 \int_0^2 \frac{3}{2}y\ dydx=\frac{3}{2}\int_0^2 y\ dy=\frac{3}{4}\Big[y^2\Big]_0^2
$$
$$
\therefore \text{work}=3
$$
