Your problem could be interpreted one of two ways: either the slant height is doubled, or the perpendicular height is doubled.
Let's look our original pyramid, call it A. I'll assume you have a square-based pyramid. Then let each side of the base of the pyramid be b units, and the slant height (not the perpendicular height h units) be s units.
Your problem could be interpreted one of two ways: either the slant height is doubled, or the perpendicular height is doubled. If it is the perpendicular height, then pyramid A has volume

, and your 'doubled' pyramid B has volume
\left( {2h} \right) = \frac{4}{3}bh = 4{V_A})
- therefore, it is four times bigger in terms of volume, and so four times heavier.
The harder way is if the slant height is doubled. Then, you end up with your volumes as
}^2}} )
and
{h_B} = \frac{{2b}}{3}\sqrt {4{s^2} - {b^2}} )
, which look a bit harder to work out, but the answer again comes out at 4. Therefore, I think your answer of 4 was correct.