So, assuming I understand your question (which I may not, I seem to have gotten a bit mixed up in your clarification of what the exact problem with the explanation was), but to the best of my ability, here is the issue. Forget for a second where the words "magnitude" and "component" were used in your conversation with the teacher or from within the problem. For the sake of this answer: magnitude is simply referring to the absolute value off any measurement, regardless of direction, and components are referring to the components of any given force. If I refer to the absolute value of a single component, I will do so by clarifying it as "the magnitude of the component". With that out of the way, here it goes:
When an object (I assume a square or other noncircular object) is sitting on an inclined plane (frame of reference being strictly two dimensional) without moving, there are (in simplified terms) three forces acting upon it. The first is the gravitational force, which is acting directly downwards (regardless of angle of the plane). The gravitational forces magnitude is equal to the mass of the object multiplied by the acceleration (so g = 9.8 m/s/s). The second force acting on the object is the normal force. The normal force is not acting directly upwards, it instead acts in a direction perpendicular to the inclined plane. The third force acting on the object is static frictional force, which is acting in the opposite direction of the possible movement, so in other words, up the ramp.
The gravitational force is split into two components that are relevant to the current situation: one component parallel to the inclined plane (and therefore parallel to the path of possible movement of the object) and the other component is perpendicular to the inclined plane (and acting in the opposite direction the normal force). I will denote these two component forces as Gp (for gravitational component parallel to the plane) and as Gq (for gravitational component perpendicular to the plane). Having defined our variables, here is the situation:
Force total = F(normal) + F(gravitational) + F(frictional)
We then split this equation into two separate considerations: one dimensional movement equation parallel to the inclined plane , and one perpendicular to the inclined plane.
Force parallel = Gp + F(frictional)
Force perpendicular = Gq + F(normal)
The component of gravity acting perpendicular to the inclined plane (Gq) cancels with the normal force. We know that the object does not rise up away from the inclined plane, nor does it sink into it. Therefore we know that, purely in terms of magnitude (absolute value), Gq = F(normal). So the perpendicular aspect of the force equations has been cancelled and is not relevant to the possible movement or stillness of the object. The only equation to concern yourself with is the parallel equation.
Force parallel = Gp + F(frictional)
This is the equation that determines the motion parallel to the plane (whether it slides up or down). Frictional forces are defined as μN, where N is normal force. This equation though, is speaking only about the absolute value, or magnitude, of the frictional force. The direction of frictional force is always opposite the direction of the possible movement. So if the parallel component of gravity (Gp) is trying to pull the object down the ramp, the frictional force acts up the ramp.
Assuming the object is not moving, and that up is positive and down is negative, we can rewrite a few of our equations....(A) = angle of inclined plane:
Force parallel = Gp + F(frictional) = -mgsin(A) + μN = 0
Force perpendicular = Gq + F(normal) = -mgcos(A) + N = 0
The two possible answers you got confused on, (c) and (d), seem to be both true. Each of the two answers are technically true, the issue is more of one with words. And I would have said this at the beginning, if it weren't for the fact that, in your explanation of the situation, it seems you have accidentally gotten confused. The frictional force's magnitude is equivalent to both the magnitude of the Gp (parallel component of gravitational force) and to the magnitude of μN. Of course, that statement is not taking vector directions into account, hence my use of the word "magnitude". Technically speaking, if direction were being included as part of the problem (without them mentioning it), then both answers would be wrong since μN could be in the same direction as N, and the direction down the ramp is obviously incorrect. The question, therefore, was referring only to the absolute numerical value, or magnitude, of the frictional force. If the problem had been written better, and the word "magnitude" had been included somewhere, it definitely would have made this a less tricky (and possibly misleading) question easier to answer. As far as I can tell, it was the wording of the question that tripped you up, and not the actual concepts behind it. Everything above is just to double and triple check your logic so you can be sure (regardless of the question) that you understand the problem.
Your teacher's explanation is true though. Whether or not it is a complete explanation aside, if the angle of the ramp is increased, and the μ stays the same, eventually the object will slip down the ramp. The math works regardless of what way you come at this (so it doesn't matter in terms of the difference between answers (c) and (d)). I hope this was helpful, my apologies for the poor formatting.