You said this is a one dimensional problem, and the drag force is proportional to the square of the velocity $v$, so
$$
F = - b v^2 = m \frac{dv}{dt} \ \ \ \Rightarrow \ \ \ \int_{v\left(0\right)}^{v\left(t\right)}\frac{dv}{v^2} = - \frac{b}{m} \int_0^t dt'
$$
The above integral (I'll let you do it, or see this) gives $v\left(t\right)$. The average velocity (or speed, since $v\left(t\right)>0$) from $t=t_1$ to $t=t_2$ is then
$$
\left<v\left(t\right)\right>_{12} = \frac{1}{t_2-t_1} \int_{t_1}^{t_2} dt \ v\left(t\right)
$$
If you had a very squishy object, it will exert a small force on the ground, whereas if you have a very hard object there will be a large force. From this it's clear that asking for the force is ambiguous, we're going to need to introduce some other variable.
As the question suggests, one thing we can do is include a variable, lets call it $\Delta t$, which tells us the duration of the collision between object and ground. The nicest way to do this is to write down the force equation you had:
$F(t) = \frac{dp}{dt}$
I've included the time on the left hand side to remind us that the force will change as a function of time over the course of the collision.What we can do is integrate this equation from $t=0$, the time of contact, to $t=\Delta t$, the time at which the object comes to rest. Then
$\int_0^{\Delta t} F(t) dt = \int_0^{\Delta t} \frac{dp}{dt} dt = \Delta p$
We can multiply and divide by $\Delta t$ to see that
$\Delta p = \Delta t \left(\frac{1}{\Delta t} \int_0^{\Delta t} F(t) dt \right) = \Delta t \,\, F_{avg,t}$
Presumably this is how you found $F = M \sqrt{2g h} / t$, but I wanted to be clear what $F$ meant in that equation.
As an alternative which doesn't use the time, we can exploit the work energy theorem:
$F(x) = \frac{dW}{dx}$
Again, integrate both sides, this time from $x=0$ to $x=\Delta x$, the total distance over which the collision occurs. This time we find
$\Delta x F_{avg,x} = W$
By the work energy theorem $W = \Delta E$ where $\Delta E$ is the change in energy of the object, so
$F_{avg,x} = \frac{1}{2\Delta x} M v^2$
This gives us a way to write down an average force without reference to the time. The tradeoff is that now we have the distance over which the collision occurs, and we find the force averaged over position rather than over time.
Best Answer
Just remember that $1 Pa = 1\dfrac{N}{m^2}$, so you just multiply by square area to obtain force. Remember that only the pressure on the end matters because the other pressure cancels out.