Given

Darlington pair:

$\beta_D=8000$

$V_{BE}=1.6\ \text{V}$

Supply voltage:

$V_{CC}=18\ \text{V}$

Base resistor:

$R_B=3.3\ \text{M}\Omega$

Emitter resistor:

$R_E=390\ \Omega$

Let base current be $I_B$.

Step 1 : Find emitter current

For Darlington pair,

$V_E=V_B-V_{BE}$

and

$V_B=18-I_BR_B$

Hence,

$V_E=18-I_BR_B-1.6$

$=16.4-I_BR_B$

Therefore,

$I_E=\dfrac{16.4-I_BR_B}{390}$

Step 2 : Use current gain relation

For Darlington transistor,

$I_E=(\beta_D+1)I_B$

$=(8000+1)I_B$

$=8001I_B$

Thus,

$8001I_B=\dfrac{16.4-3.3\times10^6 I_B}{390}$

Step 3 : Solve for $I_B$

$390(8001)I_B=16.4-3.3\times10^6 I_B$

$3.12039\times10^6 I_B=16.4-3.3\times10^6 I_B$

$6.42039\times10^6 I_B=16.4$

$I_B=\dfrac{16.4}{6.42039\times10^6}$

$I_B\approx2.55\times10^{-6}\ \text{A}$

Final Answer

$\boxed{I_B\approx2.55\ \mu\text{A}}$