Answer
The torque about the origin is $~~(56~N\cdot m)~\hat{k}$
Work Step by Step
We can write the general form of the cross-product:
$a\times b = (a_xb_y-b_xa_y)~\hat{k}$
We can express the particle's position in unit-vector notation:
$r = (3.0~m)~\hat{i}+ (8.0~m)~\hat{j}$
We can express the force in unit-vector notation:
$F = (-7.0~N)~\hat{i}$
We can find the torque about the origin:
$\tau = r \times F$
$\tau = [(3.0~m)(0)-(-7.0~N)(8.0~m)]~\hat{k}$
$\tau = (56~N\cdot m)~\hat{k}$
The torque about the origin is $~~(56~N\cdot m)~\hat{k}$
