Clearly $\Delta=\frac{1}{2} \cdot a \cdot b \sin c$
$$
\begin{aligned}
& \Rightarrow \quad \Delta=\frac{1}{2} \cdot 2 \cdot 1 \cdot \sin 60^{\circ} \\
& \Rightarrow \quad \Delta=\frac{\sqrt{3}}{2} \\
& \Rightarrow \quad \Delta^2=\frac{3}{4} \\
& \Rightarrow \quad 4 \Delta^2=3 \\
&
\end{aligned}
$$



$$
\Rightarrow \quad 4 \Delta^2=3
$$
By sine rule, we get
$$
\begin{aligned}
& \frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c} \\
\Rightarrow \quad & \frac{\sin A}{1}=\frac{\sin B}{2}=\frac{\sin 60^{\circ}}{C} \\
\Rightarrow \quad & \frac{\sin A}{\sin B}=\frac{1}{2}=\frac{\sin 60^{\circ}}{C}
\end{aligned}
$$
Considering last two terms, we get
$$
\begin{aligned}
\frac{1}{2} & =\frac{\frac{\sqrt{3}}{2}}{c} \\
\Rightarrow \quad c & =\sqrt{3}
\end{aligned}
$$
Thus, $4 \Delta^2+c^2=3+3=6$ [(Using Eqs. (i) and (iii)]