Answer
$\mu_s = 0.36$
Work Step by Step
$v_0 = (45 ~km/h)(\frac{1000 ~m}{1 ~km})(\frac{1 ~h}{3600 ~s}) = 12.5~m/s$
We can use this velocity to find the acceleration.
$a = \frac{v-v_0}{t} = \frac{0-12.5 ~m/s}{3.5 ~} = -3.57 ~m/s^2$
We can use this magnitude of deceleration to find the coefficient of static friction. We can assume that the force of static friction provides the force to decelerate the cup at this rate of deceleration:
$F_f = ma$
$mg ~\mu_s = ma$
$\mu_s = \frac{a}{g} = \frac{3.57 ~m/s^2}{9.80 ~m/s^2} = 0.36$
The coefficient of static friction $\mu_s$ is 0.36