Answer
The angular momentum about the origin is $~~0$
Work Step by Step
We can write the general form of a cross-product:
$a\times b = (a_yb_z-b_ya_z)~\hat{i}+(a_zb_x-b_za_x)~\hat{j}+(a_xb_y-b_xa_y)~\hat{k}$
We can express the particle's position in unit-vector notation:
$r = (2.0~m)~\hat{i}+ (-2.0~m)~\hat{k}$
We can express the velocity in unit-vector notation:
$v = (-5.0~m/s)~\hat{i}+(5.0~m/s)~\hat{k}$
We can find the angular momentum about the origin:
$L = m~(r \times v)$
$L = (0.25~kg)~[(0)(5.0~m/s)-(0)(-2.0~m)]~\hat{i}+[(-2.0~m)(-5.0~m/s)-(5.0~m/s)(2.0~m)]~\hat{j}+[(2.0~m)(0)-(-5.0~m/s)(0)]~\hat{k}$
$L = (0.25~kg)~[(0)~\hat{i}+(0)~\hat{j}+(0)~\hat{k}]$
$L = 0$
The angular momentum about the origin is $~~0$