Charge time capacitor in a BJT collector with a parallel resistor
\$\small V_C=V_R\$, therefore

\$\frac{1}{C}\large\int \small I_C\: dt=R(0.003-I_C)\$

Solve for \$\small I_C\$; initial condition is: \$\small t=0, I_C=0.003\: A\$.

Assuming a perfect diode that switches on at 0.7 V, the capacitor will charge linearly to 0.7 V, at which point the diode switches on and carries the 3 mA thereafter.