There are two possible sources of friction, not one - if that's not clear, you may get confused. So let's start from basics.
First, the cord can slip freely, or experience friction, sliding on the pulley. Second (and I know your question kind of said "ignore this") the pulley can turn freely or experience friction sliding on the rod that's supporting it. We will treat the two sources of friction as if they acted like one source as far as the cord is concerned, but it's important to notice it could exist and might need careful consideration.
(A third point would be the angular momentum/moment of inertia of the pulley itself, if the pulley were heavy and needed significant energy to begin to turn as the cord moves on it, but we will ignore this as well, and assume a lightweight pulley.)
I don't have drawing software here, but your answer goes like this:
Basic equation: Net Force = mass x acceleration. ($F=mA$)
Forces on the box
There are 2 forces acting on the box. A force due to gravity (call it $W$) downwards, and tension in the string (call it $T$) upwards. The box is in equilibrium so $W = T$. The force due to gravity acting on a mass $m$ is $mg$, so $W$ is easily worked out as $W = mg$. Because the box is in equilibrium, $T$, the tension in the cord, is the same as this in size, so $T = W = mg$.
Forces acting on the cord/tension in the cord
The cord (slightly simplifying as is usual for questions at this level) is also in equilibrium, so from the cord's perspective, it experiences three forces which also balance out. On one end, the force of the box, on the other end the force due to the man pulling and in the middle, any static frictional force from the contact with the pulley (that exists when the cord isn't moving). There may be some, or none. But if there is a frictional force, it's going to resist movement of the cord, so it'll act the opposite way to whichever way the cord would otherwise move.
Condition for equilibrium
Suppose the pulley can exert a force on the cord due to friction, up to an amount of $N$ newtons. Then what will happen is this:
The man pulls with force $F$. But the cord is in equilibrium. The net force from pulling and from the weight of the box is $F-W$, and because it's in equilibrium, this has to be "small enough", between $+N$ and $-N$, otherwise friction can't provide enough force to balance it and it won't stay static in equilibrium.
So remembering that $W=mg$, the condition will be that:
$$-N \leq F - m.g \leq N$$
Adding $mg$ to all terms:
$mg - N \leq F \leq mg + N$
and splitting this into separate conditions and rearranging:
$F \geq mg - N$ and $F \leq mg + N$
We can't do more because in the question, the force needed by the man to maintain equilibrium depends on 2 things - the mass of the box, and the maximum force possible due to friction, and we don't have any information to work either of these out any further.
So what this says in plain English is that the force the man has to apply, needs to be "close enough" to $mg$, that friction can supply the rest of the balancing force needed for equilibrium. If friction provided no force ($N=0$) then you'd get $F =mg$ which is the exact solution for a frictionless pulley.