From Eq. (b) of the solution of Example Analysis 4 the shaft moment T is:
T = ra\ellτ (a)
By substituting the given data values, T_{max} is:
T_{max}=\left[\left(25\right)\left(10^{-3} \right) \right]\left[\left(12.5\right)\left(10^{-3} \right) \right]\left[\left(37.5\right)\left(10^{-3} \right) \right]\left[\left(200\right)\left(10^{6} \right) \right]Nm
or
T_{max}=2.344kN m (b)