Using Stellarium, I got a sun crossing the 0° inclination at 2018-09-18 09:28:40Z, using Wikipedia's lat/long for the crater. I think it would likely be later than this, because you're in a crater, so there's some hills on the horizon.
If you're at the deepest point of the crater, you're 4.7km down, in an 85 km diameter crater, so you'd get an additional angle of...
θ = tan⁻¹ 4.7 km / ( 85/2 km ) = 6.31°
But some of this will be taken up by the Moon's curvature, so let's get its portion of the circumference
42.5 km * ( 360° / 10000 km ) = 1.53°
For some reason I feel like this is the same angle as the horizon due to some geometry but I'm not going to even try without a piece of paper, just accepting the number So depending on your altitude, you could need to get up to 4.78°, which could be as late as 2018-09-18 22:40:37Z ... though you'd see the sun on the other side of the crater first.
If you're at the deepest point of the crater, you're 4.7km down, in an 85 km diameter crater, so you'd get an additional angle of...
θ = tan⁻¹ 4.7 km / ( 85/2 km ) = 6.31°
But some of this will be taken up by the Moon's curvature, so let's get its portion of the circumference
42.5 km * ( 360° / 10000 km ) = 1.53°
For some reason I feel like this is the same angle as the horizon due to some geometry but I'm not going to even try without a piece of paper, just accepting the number So depending on your altitude, you could need to get up to 4.78°, which could be as late as 2018-09-18 22:40:37Z ... though you'd see the sun on the other side of the crater first.
"Kitto daijoubu da yo." - Sakura Kinomoto