As per the given in the question,
k×flux= 6.0v.s
Vt= 10:0 V
Ra= 1.5:00
Ia= 4:0 A
We know that rated torque (T)=k×flux×Ia
=6×4=24VsA
suppose k =1
Starting torque =KIa2 =42=16VsA
starting current (I)=RVt=1.510=6.66A
starting torque for the shunt motor = KIa= 4VsA