#### Answer

The magnitude of the angular momentum relative to the origin is $1.2~kg~m^2/s$ in the counterclockwise direction.

#### Work Step by Step

We can find the momentum vector of the particle.
$p = (m~v)~\hat{j}$
$p = (0.10~kg)(4.0~m/s)~\hat{j}$
$p = (0.40~kg~m/s)~\hat{j}$
We can express the particle's position as a vector.
$r = (3.0~m)~\hat{i}+(2.0~m/s)~\hat{j}$
We can find the angular momentum relative to the origin.
$L = r\times p$
$L = [(3.0~m)~\hat{i}+(2.0~m/s)~\hat{j}]\times [(0.40~kg~m/s)~\hat{j}]$
$L = (1.2~kg~m^2/s)~\hat{k}+0$
$L = (1.2~kg~m^2/s)~\hat{k}$
The magnitude of the angular momentum relative to the origin is $1.2~kg~m^2/s$ in the counterclockwise direction.