Answer
The magnitude of the angular momentum relative to the origin is $1.3~kg~m^2/s$ in the clockwise direction.
Work Step by Step
We can find the momentum vector of the particle as:
$p = (m~v_x)~\hat{i}+(m~v_y)~\hat{j}$
$p = (0.20~kg)(3.0~m/s)~cos(45^{\circ})~\hat{i}- (0.20~kg)(3.0~m/s)~sin(45^{\circ})~\hat{j}$
$p = (0.424~\hat{i})~kg~m/s-(0.424~\hat{j})~kg~m/s$
We can express the particle's position as a vector;
$r = (1.0~\hat{i})~m+(2.0~\hat{j})~m$
We can find the angular momentum relative to the origin.
$L = r\times p$
$L = [(1.0~\hat{i})~m+(2.0~\hat{j})~m]\times [(0.424~\hat{i})~kg~m/s-(0.424~\hat{j})~kg~m/s]$
$L = -(0.424~\hat{k})~kg~m^2/s-(0.848~\hat{k})~kg~m^2/s]$
$L = -(1.3~kg~m^2/s)~\hat{k}$
The magnitude of the angular momentum relative to the origin is $1.3~kg~m^2/s$ in the clockwise direction.
