Answer
$\tau = (50~N\cdot m)~\hat{k}$
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:
$r = (3.0~m)~\hat{i}+ (4.0~m)~\hat{j}$
We can express the force in unit-vector notation:
$F = (-8.0~N)~\hat{i}+(6.0~N)~\hat{j}$
We can find the torque about the origin on the particle:
$\tau = r \times F$
$\tau = [(4.0~m)(0)-(6.0~N)(0)]~\hat{i}+[(0)(-8.0~N)-(0)(3.0~m)]~\hat{j}+[(3.0~m)(6.0~N)-(-8.0~N)(4.0~m)]~\hat{k}$
$\tau = (50~N\cdot m)~\hat{k}$
