First of all, check your arithmetic on the X equation. You should get Vg(x) = +1.74 m/s.
Second, in the Y equation you should use the final orange speed 4.34 instead of 5.3 on the right hand side. You should get Vg(y) = -2.48 or so.
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
There is no unequivocal answer to this question.
The answer is: it depends on the 'hardness' of the bullet.
Please note that momentum is always conserved (in the absence of external forces), regardles of whether the collision is elastic or not.
Re. my first statement, let's consider the wall to be of infinite hardness, so that it undergoes no deformation during the collision whatsoever.
So only the bullet undegoes reversible deformation. Reversible because the collision is elastic. If it was inelastic, permanent deformation of the bullet would constitute energy loss, characteristic of an inelastic collision.
As a very simple model and thought experiment, let's consider the bullet to be a Hookean spring, where $F$ is the restoring force:
$$F=-kx$$
and $x$ represents the deformation of the bullet during the collision.
Assume the bullet, pre-collision, has a velocity $v$ and mass $m$, then its Kinetic energy $K$ was:
$$K=\frac12 mv^2$$
As the bullet slows down during the collision, it loses $K$, while the bullet gains elestic energy. Using Energy Conservation we can write:
$$\frac12 mv^2-\frac12 k x_{max}^2=0$$ $$\Rightarrow x_{max}=\sqrt{\frac{m}{k}}v$$ where $x_{max}$ is the maximum deformation during the collision.
We can then also write:
$$F_{max}=-k\sqrt{\frac{m}{k}}v=-v\sqrt{km}$$ where $F_{max}$ is the maximum deceleration force the bullet exerts on the wall (and vice versa of course, with $\text{N3L}$)
Because $k$ represents the 'hardness' of the bullet, it's clear $F_{max}$ depends on this hardness, as well as depending on $v$.