#### Answer

The radius of the turn is 42.5 meters.

#### Work Step by Step

Let the centripetal acceleration $a_c$ be $1.5~g$. Then, we find $r$ as:
$a_c = \frac{v^2}{r}$
$1.5~g = \frac{v^2}{r}$
$r = \frac{v^2}{1.5~g}$
$r = \frac{(25~m/s)^2}{(1.5)(9.80~m/s^2)}$
$r = 42.5~m$
The radius of the turn is 42.5 meters.