The surface gravity of a planet is very close to
$$g=frac{4pi G}{3}rho r.$$
With $g$ to be kept constant, and $frac{4pi G}{3}$ a constant, we need
$rho_Pr_P=rho_Er_E$, or
$$r_P=frac{rho_E}{rho_P}r_E,$$
with $rho_E=5.515 mbox{ g}/mbox{cm}^3$ the mean density of Earth, $r_E=6371.0 mbox{ km}$ the mean radius of Earth, $rho_P=22.59mbox{ g}/mbox{cm}^3$ the density of densest natural element osmium, and $r_P$ the radius of the fictive osmium planet.
Hence $$r_P=frac{5.515}{22.59}r_E=0.2441~r_E=1555mbox{ km}.$$
Some compression of the core of an osmium planet due to pressure is neglected.
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