#### 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 $\frac{R'}{R}$, where $R$ is the original radius of the earth.
$\frac{G~M}{(R')^2} = 3\times \frac{G~M}{R^2}$
$(\frac{R'}{R})^2 = \frac{1}{3}$
$\frac{R'}{R} = \sqrt{\frac{1}{3}}$
$\frac{R'}{R} = 0.577$
If the radius shrinks to 0.577 of the original radius, the free-fall acceleration would be three times its present value.