A spherically symmetric planet has four times the Earth's Mass and twice its radius. If a jar of peanut butter weighs 24 N on the surface of the Earth, how much would it weigh on the surface of this planet

Close to the surface of the Earth, the weight of an object is just the magnitude of the force that gravity exerts on the object,i.e., the product of the mass of the object times the acceleration due to gravity, 9.8 m/s2.
This force can be calculated using the Newton's Universal Law of Gravitation that states that two masses attracts each other with a force that is proportional to the product of their masses and inversely proportional to the square of the distance between them.
Mathematically, we can write this law as follows:

[tex]F_{g} = \frac{G*m_{1}*m_{2}}{r_{12}^{2} }[/tex]

[tex]F_{g} = \frac{G*m_{E}*m_{j}}{r_{E}^{2} } = m_{j} * g = 24 N (1)[/tex]

Now, if we put the same jar on the surface of a planet with four times the mass of the Earth, and twice its radius, we can apply the same expression (1) replacing mE by 4*mE, and rE by 2rE, as follows:

[tex]F_{gp} = \frac{G*4*m_{E}*m_{j}}{(2*r_{E})^{2} } =\frac{G*m_{E}*m_{j}}{r_{E}^{2} } = F_{g}[/tex]

Since Fgp = Fg, this means that the weight of the jar would be the same than on the surface of the Earth, i.e., 24 N.