Let's run a few numbers. We'll assume that 10% of the energy from the explosion is radiated as visible light -- that's a good enough estimate for this purpose. 10% of 200 megatons is 84,000,000,000,000,000 joules, or 8.4E16. Figure it lasts for ten seconds, and that's 8.4E15 watts.
The Moon (and Earth) get about 1,360 watts/sq.m of solar radiation. Let's assume 50% of that is visible light, 680 watts/sq.m. The Moon's albedo is 0.12, so 12% of that gets reflected, 82 watts/sq.m. We'll call it 84, or 8.4E1, because it makes the sums ever so much easier.
So, the bomb will emit about as much light as 1E14 sq.m of lunar surface. The radius of the Moon is 1,700,000 metres, so the area of the disc of the full Moon is 9E12 square metres. Call it 1E13.
Therefore, for a few seconds the bomb will be something like 10 times as bright as the full Moon, and will thus be easily visible, but not immensely dramatic.
There won't be any dust clouds (for more than a minute or two), because there's no atmosphere to hold them up. The dust will fall back to the surface as quickly as any larger ejecta.
There would be no long-term effects other than a small (not visible without a pretty good telescope) new crater. There have been many thousands of bigger meteorite impacts on the Moon.