#### Answer

If the radius shrinks to 0.577 of the original radius, the free-fall acceleration would be three times its present value.

#### Work Step by Step

Let $R'$ be the radius of the earth after it shrinks. Note that the mass $M$ of the earth does not change. We can find the value of $R'$ in terms of $R$, where $R$ is the original radius of the earth.
$\frac{G~M}{(R')^2} = 3\times \frac{G~M}{R^2}$
$(R')^2 = \frac{1}{3}~R^2$
$R' = \sqrt{\frac{1}{3}}~R$
$R' = 0.577~R$
If the radius shrinks to 0.577 of the original radius, the free-fall acceleration would be three times its present value.