Solution:
The smaller area enclosed by the circle, $x^{2}+y^{2}=4$, and the line, $x+y=2$, is represented by the shaded area ACBA as

It is observed that,
$\text { Area } A C B A=\text { Area } O A C B O-\text { Area }(\triangle O A B)$
$\begin{array}{l}
=\int_{0}^{2} \sqrt{4-x^{2}} d x-\int_{0}^{2}(2-x) d x \\
=\left[\frac{x}{2} \sqrt{4-x^{2}}+\frac{4}{2} \sin ^{-1} \frac{x}{2}\right]_{0}^{2}-\left[2 x-\frac{x^{2}}{2}\right]_{0}^{2}
\end{array}$
$=\left[2 \cdot \frac{\pi}{2}\right]-[4-2]$
$=(\pi-2)$ units
As a result, the coreect answer is B.