$$
\begin{array}{l}
\text { 6. Let } a \in \mathbf{R}^{+}, \\
A=\left\{(x, y) \left\lvert\,(x-1)^{2}+(y-2)^{2} \leqslant \frac{4}{5}\right.\right\} \\
\text { and } B=\{(x, y)|| x-1|+2| y-2 \mid \leqslant a\}
\end{array}
$$

are point sets in the plane $x O y$. Then $A \subseteq B$ if
(A) $a \geqslant 2$
(B) $a \geqslant \sqrt{5}$
(C) $a \geqslant \sqrt{6}$
(D) $a \geqslant 3$
6. (A).

Set $A$ is a circular region with center at $(1,2)$ and radius $\frac{2}{\sqrt{5}}$. Set $B$ is a rhombus with the diagonal intersection at $(1,2)$, and the two diagonals are parallel to the coordinate axes, with the diagonal parallel to the $x$-axis being twice the length of the other diagonal.
$A \subseteq B$, if and only if the circle is inside the rhombus and tangent to it, which is the minimum case for the value of $a$ (i.e., the length of $EA$). At this time,
$$
\begin{aligned}
a & \cdot \frac{a}{2}=r \cdot \sqrt{a^{2}+\left(\frac{a}{2}\right)^{2}} \\
& =\frac{2}{\sqrt{5}} \cdot a \cdot \sqrt{\frac{5}{4}} . \\
\therefore a_{\text {min }} & =2, a \geqslant 2 .
\end{aligned}
$$