What light looks like from the surface
The sun in your world is half-bright, half-dark. Assume that it is a perfect sphere with any radius greater than zero. Therefore, for any point on the surface, the percentage of the sun that is facing that point at any given time is
$$\cos^2\phi\cos^22\theta + \frac{1}{2}\sin^2\phi.$$
Here are three plots of what that looks like at the equator:
At a point in the temperate zone at about 40 degrees latitude:
And at the poles:
How to measure latitude between two points
Without sunrise over a horizon or a pole star, it is much harder to measure latitude. The best you can do is to measure the brightness of the sun at various points in the day, and to compare with the trigonometeric properties of its brightness from the last section.
The the best of my knowledge, there is no way to scientifically measure brightness until you have a photographic plate. That will not be available to people of Renaissance technology. However, you can at the very least dead-eye reckon brightness, so we will assume some sort of brightness metric. One potential way to measure brightness is that brightness is proportional to the 'bright' portion of the sphere of the sun that is visible to you from your location. If you look at the sun through a dark lens, you may be able to measure this, depending on how large the radius of the sun is.
In this case, you can calculate latitude relative to the equator easily. Maximum brightness at latitude $\phi$ ($\max(B_\phi)$) is $1 - 1/2\sin^2\phi$ times that at the equator ($\max(B_{eq})$), solve backwards for $\phi$:
$$\phi = \sin^{-1}\sqrt{2\left(1 - \frac{\max(B_\phi)}{\max(B_{eq})}\right)}.$$
It is possible that maximum brightness at the equator is a well known standard value in your scientific community, even for those who have never been there. You have to know, or at least be able to estimate, the maximum brightness to be able to calculate latitude. As an alternate measurement, not2 that equatorial max brightness is twice the constant brightness at the poles. You can re-work everything in terms of that value as
$$\max(B_{eq}) = 2B_{pole}.$$
Now there are two ways to measure radius
Assuming the hollow world is a perfect sphere, there are two ways to measure the radius. you can compare latitude of two points to the distance between them, or you can compare the time it takes the terminator to travel between two points.
The terminator line
The terminator is important here, because with no stars in the background, the only way to ensure that two points are at exactly the same longitude is to visually signal when the terminator passes by the points. This can be done using one lighthouse to signal another point within visual range of that lighthouse. Since your world is hollow and thus concave from the point of view of someone on the surface, the line of sight of a lighthouse is actually very long, limited only by atmospheric attenuation of light (due to water vapor, or whatever).
The terminator can be exactly identified by looking at the sun through a telescope with a dark lens. As soon as there is no bright patch visible, the terminator line has passed.
Comparing latitude method
The latitude way is take two points that you know are at the same longitude, and calculate their latitude using the above method and distance ($d$) between them, using what ever method.
If the latitude delta is $\alpha$, then the polar radius of the hollow Earth is $$r_{pole} = d/\alpha$$ in radians.
Timing the terminator method
To time the terminator, you will need to get two points at the same latitude (confirmed using the methods above), and measure the distance ($d$) between them and the time it takes the terminator to pass between them ($t$). You also need to know the length of the day $t_{day}$.
Then, the equatorial radius of the hollow Earth is $$r_{eq} = 2\pi d\frac{ t_{day}}{t}.$$ Note this only works if the points are less than $\pi$ radians apart on the surface, that is, in the same hemisphere.
These numbers might not be the same!
These two methods could give you different answers, polar and equatorial radius. If you have a perfect sphere, the two calculated radii should be the same, but if your planet is inscribed within an oblate spheroid with an equatorial bulge(like our Earth), or even a polar bulge, then the two numbers will not be the same.
Conclusion
It is actually pretty difficult to time astronomical pheomena with only one available astronomical object. But, you can use the unique configuration of the sun's surface to do this.
This does require some photometric skills more advanced than what was available in the Renaissance, which I'm not sure how to replicate with Renaissance technology, but I have faith that the Galileo's and Newton's of the world could figure it out. Also, there needs to be some sort of standard measurement of equatorial brightness to make these calculations, but given this number's importance in observational 'heliometry,' I would expect this number to be a well established topic in a Hollow Earth's scientific community.
