What is the maximum voltage that can be applied to the base of a bias resistor built-in transistor (BRT)? (How many watts is the allowable power dissipation of the built-in resistors?)

You do not need to consider the maximum allowable base voltage for typical applications (in which input voltage (V_I) is lower than 10 V), but care is required when V_I is higher than 10 V.
The maximum input voltage (V_I) is mainly determined by the following factors:
1.    Value of the series base resistor (R₁) (i.e., the power dissipated by R₁)
2.    Maximum collector current (I_C)
The maximum input voltage varies with the usage conditions of a BRT (e.g., hFE), but it is generally determined by the collector current when the R1 value is small and by the R1 value when it is large.

1. Relationship between the value of the series base resistor (R₁) and the input voltage (V_I)
The allowable power dissipation of the built-in resistors of a BRT is 1/8 W. Let the base-emitter voltage of the internal transistor in the “on” state be V_be. Then, the current flowing through R₁ (I_B) is expressed as follows. For the sake of simplicity, let V_be = 0.7 V.
I_B ＝ ( V_I – V_be ) / R₁ = ( V_I – 0.7 ) / R₁
The power dissipation of R₁ due to I_B must not exceed 1/8 W. Hence:
    1/8 W > R₁ * { ( V_I – 0.7 ) / R₁ }^2
              > ( V_I – 0.7 ) ^2 / R₁    
    V_I < √( R₁ / 8 ) + 0.7

2. Relationship between the maximum collector current (I_C(max)) and the input voltage (V_I)
The maximum collector current (I_C(max)) is specified as an absolute maximum rating, which must not be exceeded even instantaneously.
Since the BRT operates in the saturation region, its h_FE is in the range of 10 to 20. The current flowing to the base (b) of the internal transistor (I_b) can be calculated by subtracting the current flowing to R₂ (I_R2) from the I_B value calculated above:
    I_b = I_B – I_R2 = ( V_I – V_be ) / R₁ – V_be / R₂ = ( V_I – 0.7 ) / R₁ – 0.7 / R₂
Since the collector current (I_C) is equal to I_b times h_FE, the following equation must be satisfied:
    I_C (max) > I_C = h_FE * I_B = h_FE * {( V_I – 0.7 ) / R₁ – 0.7 / R₂ }
    V_I < R₁ * I_C(max) / h_FE +( R₁ + R₂ ) * 0.7 / R₂
The lower of the V_I values calculated based on the above two factors is the maximum allowable base voltage.
1. Value of the series base resistor (R₁)
    V_I < √( R₁ / 8 ) + 0.7
2. Maximum collector current (I_C(max))
    V_I < R₁ * I_C (max) / h_FE +( R₁ + R₂ ) * 0.7 / R₂
Figure 2 and Figure 3 show the relationships between input voltage and collector current at different resistor values.
Although the maximum input voltage and its limiting factors depend on the h_FE (I_C/I_B) of the BRT, the following are true:

• When the R₁ value is small: The maximum input voltage is determined by the absolute maximum collector current specified in the datasheet.
• When the R₁ value is large: The maximum input voltage is determined by the power dissipation of R₁.