Answer
$$
\begin{aligned}
& \mathbf{M}_B=\{-3.36 \mathbf{k}\} \mathrm{N} \cdot \mathrm{m} \\
& \alpha=90^{\circ} \\
& \beta=90^{\circ} \\
& \gamma=180^{\circ}
\end{aligned}
$$
Work Step by Step
$$
\begin{aligned}
& F=20\left(\sin 60^{\circ} \mathbf{i}-\cos 60^{\circ} \mathbf{j}\right)=\{17.32 \mathbf{i}-10 \mathbf{j}\} \mathrm{N} \\
& \mathbf{r}_{B A}=\{-0.01 \mathbf{i}+0.2 \mathbf{j}\} \mathrm{m}
\end{aligned}
$$
Moment of Force $\boldsymbol{F}$ about point $\boldsymbol{B}$.
$$
\begin{aligned}
\mathbf{M}_B & =\mathbf{r}_{B A} \times F \\
& =\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
-0.01 & 0.2 & 0 \\
17.32 & -10 & 0
\end{array}\right| \\
& =\{-3.3641 \mathbf{k}\} \mathrm{N} \cdot \mathrm{m} \\
& =\{-3.36 \mathbf{k}\} \mathrm{N} \cdot \mathrm{m}
\end{aligned}
$$
Now the unit vector for $\mathbf{M}_B$ is $\mathbf{u}=-\mathbf{k}$. Thus, the coordinate directi angles of $\mathbf{M}_B$ are
$$
\begin{aligned}
& \alpha=\cos ^{-1} 0=90^{\circ} \\
& \beta=\cos ^{-1} 0=90^{\circ} \\
& \gamma=\cos ^{-1}(-1)=108^{\circ}
\end{aligned}
$$
