A car is negotiating a flat, unbanked circular turn. For a turn of...

The maximum safe speed for a car on a flat circular track is determined by the static friction providing the necessary centripetal force.

Equating the maximum static friction to the required centripetal force:
$\dfrac{mv^2}{R} = \mu mg$

Solving for $v$, we get:
$v = \sqrt{\mu R g}$

This shows that the maximum safe speed is directly proportional to the square root of the radius of the turn ($v \propto \sqrt{R}$).

When the radius is increased to $4R$, the new maximum safe speed $v'$ is:
$v' = \sqrt{\mu (4R) g} = 2\sqrt{\mu R g} = 2v$.

Answer: $2v$