Q:

How to calculate density of unknown planet?

Having approached an unknown planet, a 
spaceship went in a low circular orbit. Would 
astronauts be able to determine the average 
density of the planets using only a watch

A:

Planets are usually rotating, and so you can't time two passes over a single geographical feature to time an orbit. But you can time successive sunrises to get a good approximation of the orbital period, or some other astronomical coordination between the planet's limb and the celestial sphere.
From this answer:

Starting with
$$T = 2\pi\sqrt{\frac{a^3}{\mu}} = 2\pi\sqrt{\frac{a^3}{Gm}}$$
and
$$\rho = \frac{m}{\frac{4}{3} \pi R^3}$$
$$\frac{1}{m} = \frac{1}{\frac{4}{3} \rho \pi R^3}$$
where $\rho$ is the density and $R$ is the body's radius. Then:
$$T = 2\pi\sqrt{\frac{a^3}{G\frac{4}{3} \rho \pi R^3}},$$
and if you set $a=R$, you get:
$$T = 2\pi\sqrt{\frac{3}{4G\pi \rho}},$$
and finally
$$T = \sqrt{\frac{3 \pi}{G\rho}}.$$

And finally, flipping that around, we get:
$$\rho = \frac{3 \pi} {G T^2}.$$
The catch is the step where you set the radius of the circular orbit equal to the radius of the planet; $a=R$.
On an airless world you can get pretty close for a little while, but bodies are usually nonuniform (lumpy) and so an orbit that skims near the surface will quickly get perturbed and crash.
So to do the experiment safely you need to orbit a safe distance above the surface, and this makes $a=R$ a source of error.
One solution is to use a hand-held laser altimeter (if such a thing exists) but probably that won't be found in an astronaut's watch, not even a fancy Scott Kelly watch.
With a conventional watch, one way to estimate altitude would be to measure the time of more celestial events, like a sunrise plus a sunset, as long as you can work out the 3D geometry to convert that to an altitude. The higher the altitude, the longer the day and shorter the night is.
Once you have altitude $h$, you can solve for both the radii of the planet $R$ and of the orbit $a$.

