How dense would planet earth have to be to have the same gravity as Jupiter?
There are at least two interpretations to this problem:

Per Wikipedia, Jupiter's surface gravity is $2.528$ times Earth's. Thus, 
if the Earth were $2.528$ times denser, it would have the same surface 
gravity as Jupiter. The Earth's current density is $5.514$ grams per 
cubic centimeter, so the new density would be $2.528 \times 5.514$, or about 
$13.9394$ grams per cubic centimeter. This assumes we change Earth's mass, but not its radius.

@Rob_Jeffries answer assumes the Earth's mass remains constant and the radius changes. If the radius shrinks by a factor of $2$, the volume decreases by a factor of $8$, and the planet becomes 8 times more dense. The surface gravity increases by $4$, since it depends on the radius squared. In general, shrinking the planet's radius by $x$ will increase the density by $x^3$ and the gravity by $x^2$. If we want the gravity $2.528$ times higher, we choose $x = \sqrt{2.528}$ or right around $1.590$. This makes $x^3$ equal to about $4.019$. Multiplying that by Earth's current density of $5.514$, we get $22.16$ grams per cubic centimeter, pretty close to what Rob got.

So, you can't really change the density without changing anything else: either the mass or the volume must change.