Answer
$\tau = (-90t^2~N\cdot m)~\hat{k}$
Work Step by Step
Note that a line can be drawn from the point $(2.0~m, 5.0~m, 0)$ to meet the x axis at a right angle and the length of this line is $5.0~m$ in the -y direction.
We can find the magnitude of the angular momentum about the point $(2.0~m, 5.0~m, 0)$:
$L = r_{\perp}~mv$
$L = (5.0~m)~(3.0~kg)(2.0t^3~m/s)$
$L = 30t^3~kg~m^2/s$
Note that the velocity is in the -x direction. By the right hand rule, the direction of $r \times v$ is in the -z direction.
Therefore:
$L = (-30t^3~kg~m^2/s)~\hat{k}$
We can find the torque about the point $(2.0~m, 5.0~m, 0)$:
$\tau = \frac{dL}{dt}$
$\tau = (-90t^2~N\cdot m)~\hat{k}$
