Show that the torque for x = b is $M_y -b\rho S$ 
The area R on the xy-plane corresponds to a thin metal plate with the
  area S and a constant density $\rho$. $M_y$ is the plate's moment
  corresponding to the y-axis.
a) Show that the moment corresponding to $x = b$ is $M_y - b\rho S$,
  if the plate is right from $x = b$.
b) Show that the moment corresponding to $x = b$ is $b\rho S - M_y$,
  if the plate is left from $x = b$.

Now, I don't actually know much about physics. I assume they are talking about this moment, but I can't be sure. I also assume the answer has something to do with the fact that $m = \rho \cdot V$, but otherwise I'm stuck.

Saying that $M_y$ is the moment corresponding to the $y$-axis (i.e., the line $x = 0$ is saying (according to the definition linked in Sou's comment) that 
$$
M_y = \iint_R x^2 \rho~dy~dx
$$
where here $\rho$ is the density, not the "distance to the axis" as it is in the linked wikipedia page. 
The thing you're supposed to compute replaces $x$ (the distance from a point $(x, y)$ to the $x$ axis) with $x - b$ (the distance from $(x, y)$ to the line $x = b$), i.e., you're computing
$$
M' = \iint_R (x-b)^2 \rho~dy~dx
$$
"Where the plate is right from $x = b$" means that $x > b$ for all points $(x, y)$ of the plate; presumably this helps somehow in the simplifications you have to make. 
Anyhow, do some algebra to relate $M'$ to the known value $M_y$, and see where it gets you. 
