Answer
$\approx 8.9\;mm$
Work Step by Step
The Schwarzschild radius is given on page 963: $\frac{2GM}{c^2}$. Put in values for the Earth.
$$R_{S,E}=\frac{2GM}{c^2}$$
$$R_{S,E}=\frac{2(6.67\times10^{-11}Nm^2/kg^2)( 5.98\times10^{24}kg)}{(3.00\times10^8m/s)^2}$$
$$\approx 8.9\;mm$$