Answer
The torque ${\bf{\tau}}$ about $O$:
${\bf{\tau}} = 204.79{\bf{k}}$ N-m
Work Step by Step
From Figure 22(A) we see that the angle between the position vector ${\bf{r}}$ and the force ${\bf{F}}$ is $125^\circ $. Recall that the torque ${\bf{\tau}}$ about $O$ is defined by ${\bf{\tau}} = {\bf{r}} \times {\bf{F}}$.
Let $||\tau ||$ denote the magnitude of ${\bf{\tau}}$. So,
$||\tau || = ||{\bf{r}}||||{\bf{F}}||\sin \theta $,
where $\theta$ is the angle between ${\bf{\tau}}$ and ${\bf{F}}$. In this case: $\theta = 125^\circ $.
We have the length of the arm: $||{\bf{r}}|| = 10$ and the magnitude of the force: $||{\bf{F}}|| = 25$. So,
$||\tau || = 10\cdot25\cdot \sin125^\circ \simeq 204.79$ N-m
Since ${\bf{r}}$ and ${\bf{F}}$ lie in the $xy$-plane, by the right-hand rule, the torque vector ${\bf{\tau}}$ points in the direction of positive $z$-axis. So,
${\bf{\tau}} = 204.79{\bf{k}}$ N-m
