I need some help with a firepower calc- assuming the planet in the pic is earth-sized...?

The best you will get is a rough approximation, but let me see if I can help.

First you need to do some geometric construction. Draw two lines that are tangent to the planet's surface in two different places. Then draw two lines perpendicular of the tangent lines where they meet the planet surface and go inward. Where the two perpendicular lines cross will be the planet's center, and if the planet is sufficiently spherical in shape, this will give you a scale since the perpendicular lines will represent the radius of the planet (which is earth-sized).

Using this as a measuring stick, you can get a general approximation of the radius of the detonation (measure it at its widest point since you are not looking straight down on it).

Now you have another problem. Was the picture taken early in the explosion, in the middle or near the end? If near the end, then you have some basis by which you can estimate the explosive energy. If near the beginning, or in the middle, then you really don't know how large the cloud actually got so you have nothing to go by.

As for destructive power, the best guess would be to compare existing measurments to the ones you come up with in your analysis. The Trinity test was about 19 kilotons and had a fireball about 300 feet wide. The Sedan test was about 104 kilotons and had a fireball 1280 feet wide. The destructive power of a Mark V warhead (Titan II ICBM) was said to be about 9.7 megatons and would gouge out a crater a little more than half a mile wide. (That is the size of the fireball, not the radius of destruction.) You can interpolate from that if you wish, but also bear in mind that while the fireball size will be more or less the same regardless, the radius of destruction for a ground burst is much smaller than that of an air burst.

Good luck. I'd be interested in knowing what you come up with.