2 years 9 months, 5 existing answers, and no one has actually answered the very simple question that was actually asked.
Ignore how such a structure is made stable. If you actually flattened out the Earth like this, what would the gravity be like?
At the center of the disk and near the surface--the "North Pole"--the gravitational environment is well-approximated as that of an infinite plane. An infinite plane produces a uniform gravitational field above it whose strength is directly proportional to the areal density of mass below. So, we just take the area of the Earth and divide by the mass--we don't even need to bother figuring out the exact dimenions of the resulting cylinder, and then multiply by $2\pi G$, and we get an answer of $~4.91m/s^2$--so, about half a g. Which makes sense, because, if you flatten out the Earth like that, you are moving most of the mass farther away from the North Pole than where it started, so the force should get weaker.
Now, this plane is of course not actually infinite, so as you move a significant distance away from the North Pole, corrections from the edge conditions will become significant--high above the North Pole, gravity will eventually start to decrease even further, and as you move along the surface away from the North Pole gravity will both decrease in strength and start tilting so it feels like you are walking uphill. That's what will force the structure to collapse back into a sphere, unless you arrange for a counter-force. Turns out just spinning the disk is not sufficient--the inward gravitational force changes according to a complicated curve that is not matched by the linear increase in centrifugal force. You can imagine supermaterials, like Larry Niven's scrith, that rigidly hold the structure in the flat disk shape against gravity, but if you accept a slight modification, you can get a stable structure by spinning the disk and letting it thin towards the edges--so, it's actually a highly oblate spheroid with an extremely short day, in which case gravity always remains perpendicular to the ground, ends up slightly higher than the uniform plane estimate at the North Pole, and smoothly decreases to whatever lower limit you want based on the chosen spin rate--down to zero, where the equator is effectively in orbit--as you get towards the edge.