If our sun was compressed down to a radius of 2 km, how much would a spoonful of the sun weigh on earth in pounds?

Radius of the Sun . . . 695,700 kilometers

Mass of the Sun . . .  1.989 x 10³⁰ kilograms

Volume of the Sun = (4/3) (π R³) = (4/3 π) (6.957 x 10¹⁰ cm)³

= 1.41 x 10³³ cm³

Average density of the sun = (mass)/(volume) =

(1.989 x 10³⁰ kg) / (1.41 x 10³³ cm³) = 1.41 gram/cm³

This already smells incorrect to me, but I'm plunging on nonetheless.

Now you want to squeeze the sun down to (2/695,700) =  2.87 x 10⁻⁶ of its present radius.  So its density will increase by the reciprocal of that same factor, and become

(1.41 g/cm³) / (2.87 x 10⁻⁶) = 4.913 x 10⁵ g/cm³

The stench of wrongness is becoming pervasive now, but I've gone too far to either back out OR go back and look for the mistake.  I have no choice but to just keep going.

1 teaspoon = 4.93 cm³

It holds  (4.93cm³ x 4.913 x 10⁵ g/cm³) = 2,422 kilograms

1 Tablespoon = 3 teaspoons = 14.79 cm³

It holds (14.79cm³ x 4.913 x 10⁵ g/cm³) = 7,266 kilograms

If there were some space probe scheduled to launch tomorrow and relying on these calculations, I would certainly start again at the top, go through everything again, and perhaps carry things out to another 2 or 3 decimal places.  But you know what ?  I have actual work to do, and I see that the only thing riding on this is 5 points, so I'd say that at this point, what I've done is what we've got.  If you ever do find out the correct answer, I'd be curious to know if I'm anywhere in the ballpark.  Other than that, you're welcome.

The magnetic poles of Earth are at the same locations of as the geographic poles is false.

Explanation: