If transporting the stones dry, the minimum volume of wood required:

$(V_{stone}\rho_{stone} + V_{wood}\rho_{wood})g = V_{wood}\rho_{water}g$
$\therefore V_{wood}(\rho_{water} - \rho_{wood}) = V_{stone}\rho_{stone}$
$\therefore V_{wood} = \frac{V_{stone}\rho_{stone}}{\rho_{water} - \rho_{wood}} = 13.05m^3$
$V_{wood}/(\pi r_{tree}^2) = length_{tree} = 415.4m$

That's nearly half a kilometer of sizeable trees!

If the stones could be transported wet, which would of course require a river about 2 feet deeper, then less wood would have been required:

$(V_{stone}\rho_{stone} + V_{wood}\rho_{wood})g = (V_{wood} + V_{stone})\rho_{water}g$
$\therefore V_{wood}(\rho_{water} - \rho_{wood}) = V_{stone}\rho_{stone}$
$\therefore V_{wood} = \frac{V_{stone}(\rho_{stone} - \rho_{water})}{\rho_{water} - \rho_{wood}} = 8.55m^3$
$V_{wood}/(\pi r_{tree}^2) = length_{tree} = 272.2m$

That's still a lot of trees, but considerably fewer, maybe 25 large trees. That is the absolute minimum value though, at which the object will have neutral buoyancy, i.e. will have the overall density of water.