Answer
$\tau = (6.0~N\cdot m)~\hat{j}+(8.0~N\cdot m)~\hat{k}$
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:
$r = (-4.0~m)~\hat{j}+ (3.0~m)~\hat{k}$
We can express the force in unit-vector notation:
$F =(2.0~N)~\hat{i}$
We can find the torque about the origin on the particle:
$\tau = r \times F$
$\tau = (0)~\hat{i}+[(3.0~m)(2.0~N)-0]~\hat{j}+[0-(2.0~N)(-4.0~m)]~\hat{k}$
$\tau = (6.0~N\cdot m)~\hat{j}+(8.0~N\cdot m)~\hat{k}$
