Hotter than the sun

Okay, that was a little weird.

Edit: Here's how it's worked out

$M=2\times { 10 }^{ 30 } \times kg$
$R=7\times { 10 }^{ 8 } \times meter$
$T=5780 \times kelvin$
$Calorie=1000(0.00116222) \times watt \times hour$
$\sigma =5.670373\times { 10 }^{ -8 }watt \times{ meter }^{ -2 } \times{ kelvin }^{ -4 }$

$\sigma$ is the Stefan-Boltzman constant

Total radiated energy from sun in $24$ hour period divided by sun’s mass

$Sun=\dfrac { 24 \times hour }{ M } \left( 4\pi { R }^{ 2 }\sigma { T }^{ 4 } \right)$

Total radiated energy from body in $24$ hour period divided by body’s mass

$Body=\dfrac { 2000 \times Calorie }{ 75 \times kg }$

Ratio of Body to Sun

$\dfrac { Body }{ Sun } =6627.46...$

which is a dimensionless ratio, i.e., all the physical dimensions drop out

Note: Calorie, with capital "C", is $1000$ times greater than calorie, with lower case "c". Calories is commonly used for foods, while calories is used for stuff like locomotive steam engine thermodynamic analysis. Maybe that explains why that family is glowing so much.

$P = A \sigma T^4$

$A$ = $4 \pi R^2$ = $4 \pi (7 \times 10^8)^2 = 6.15752160103599 \times 10^{18} m^2$

$T^4$ = $5780^4$ = $1116121190560000 K^4$

$P = 6.15752160103599 \times 10^{18}\times 5.7 \times 10^{-8} \times 1116121190560000$ = $3.89698671867486 \times 10^{26} W.$

$\displaystyle \frac{3.89698671867486 \times 10^{26} W}{2 \times 10^{30} kg}$ = $1.94849335933743 \times 10^{-4} W/ kg$

Since 2000 Calories = 2000 $\times$ 4184 Joules = 8368000 J,

$\displaystyle \frac{8368000}{24 \times 3600} = 96.8518518518519 W;$

$\displaystyle \frac{96.8518518518519 W}{75 kg} = 1.29135802469136 W/ kg$

Ratio = $\displaystyle \frac{1.29135802469136}{1.94849335933743 \times 10^{-4}}$ = $6.59302174169231E+3$

With $5.670373$ $\times 10^{-8}$ instead of $5.7$ $\times 10^{-8}$ , $6.62746946764281E+3$ is obtainable.

This ratio may give reasonable fact of why the sun can last for a long period of time in space.

Answer: $\boxed{6593}$ $else$ $\boxed{6627}$