$\text{A.}$ $\frac{3}{5} \sqrt{10}$ $\text{B.}$ $\frac{3}{10} \sqrt{10}$ $\text{C.}$ $\sqrt{10}$ $\text{D.}$ $2$

A

#### 解析：

【详解】解: 过点 $D$ 作 $D H \perp A F$ 于点 $H$,
\begin{aligned} & \because \angle A B C=45^{\circ}, A D \perp B C, \\ & \therefore A D=B D, \\ & \because \tan \angle A C B=\frac{A D}{C D}=3, \end{aligned}

\begin{aligned} & \therefore A D=3 x, \\ & \therefore B C=3 x+x=4, \\ & \therefore x=1, \\ & \therefore C D=1, A D=3, \end{aligned}

$\because$ 将 $\triangle A D C$ 绕点 $D$ 逆时针方向旋转得到 $\triangle F D E$,
\begin{aligned} & \therefore D C=D E, D A=D F=3, \angle C D E=\angle A D F, \\ & \therefore C E C \sim D A F, \\ & \therefore \angle D C E=\angle D A F, \\ & \therefore \tan \angle D A H=3, \end{aligned}

\begin{aligned} & \text { 设 } A H=a, D H=3 a, \\ & \because A H^2+D H^2=A D^2, \\ & \therefore a^2+(3 a)^2=3^2, \\ & \therefore a=\frac{3 \sqrt{10}}{10}, \\ & \therefore A H=\frac{3 \sqrt{10}}{10}, \\ & \because D A=D F, D H \perp A F, \\ & \therefore A F=2 A H=\frac{3 \sqrt{10}}{5}, \text { 故 A 正确. } \end{aligned}