Update: I can finally answer to your question!!
You see, stars are pretty tricky to work with. I don't know how the nuclear reactions would happen in such small star because they'll likely never happen, I'm not quite at atom science.
So, we will do a
trick: we'll measure the energy of the 'star' through the radiation energy, ignoring nuclear reactions. Let's say, the energy of a newborn 'star'. Its birth is the ionization of its own mass through a high enough temperature.
The Sun's most dominating
material is
hydroge (
~73%), which has the 1st (and only) ionization energy at
13.59844 eV, which equals
157803.1649722 K (
Holy crap!!!). Note that
1 eV is
11604.505 K.
Now let's calculate the
radiation energy of such star! We need temperature
T (Known as
157803.1649722 K), emissivity
╬Á (Tricky one! I'm noob at hydrogen plasma emissivity, so I'd highballing it at
1), surface area
A (varies) and the Stefan-Boltzmann Constant
¤â (
5.67 * 10-8 W/m2*K4).
The formula for the
radiation energy calculation is
P = ╬ÁA¤âT4, where
P is measured in
watts (J/s),
A is measured in
square meters,
¤â is measured in
W/m2*K4 and
T is measured in
Kelvins.
The formula for the
surface area is
A = ¤Çd2, where
A is measured in
square meters and
d is measured in
meters.
Let's go!...
A baseball-sized star's power is
W = ╬Á¤Çd2¤âT4 = 1 * 3.14159 * 0.073662 meters * (5.67 * 10-8 W/m2*K4) * 157803.16497224 Kelvins = 599320083379.7869 W (J/s), which would equal to
8-A, Multi-City Block level tier.
A basketball-sized star's power is
W = ╬Á¤Çd2¤âT4 = 1 * 3.14159 * 0.2405382 meters * (5.67 * 10-8 W/m2*K4) * 157803.16497224 Kelvins = 6390911351435.709 W (J/s), which would equal to
Low 7-C, Small Tow tier.
A cannonball-sized star's power is
W = ╬Á¤Çd2¤âT4 = 1 * 3.14159 * 0.82 meters * (5.67 * 10-8 W/m2*K4) * 157803.16497224 Kelvins = 7.069283136340*1013 W (J/s), which would equal to
7-C, Town level tier. (Tsar Cannon is a shotgun, not a cannon, which is a common misconception)