On an unbanked road, frictional force alone can provide the centripetal force needed to keep the cyclist moving on a circular turn without slipping. If the speed is too large, or if the turn is too sharp (i.e. of too small a radius) or both, the frictional force is not sufficient to provide the necessary centripetal force, and the cyclist slips. The condition for the cyclist not to slip is given by
v 
2
 ≤μ 
s
​
 Rg
Now, R=3 m,g=9.8 m s 
−2
 ,μ 
s
​
 =0.1. That is, μ 
s
​
 Rg=2.94 m 
2
  s 
−2
 ,v=18 km/h=5 m s 
−1
 ; i.e., v 
2
 =25 m 
2
  s 
−2
 . The condition is not obeyed. The cyclist will slip while taking the circular turn.
 The word "slip" should not appear in your response.
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