Before we even begin it would be very helpful to draw out a simple layout of the circuit. Then we go ahead and apply kirchoffs second law(sum of voltages around a loop must be zero) on the circuit and we obtain the following differential equation,

$-V +Ldi/dt+Ri=0$

where V is the electromotive force applied to the LR series circuit, Ldi/dt is the voltage drop across the inductor and Ri is the voltage drop across the resistor. we can re write the equation as,

$di/dt+Ri/L=V/L$

Then we first solve for the homogeneous part given by,

$di/dt+Ri/L=0$

we obtain,

$i(t)_{h} =I_{max}e^{-Rt/L}$

This is only the solution to the homogeneous part, The final solution would be given by,

$i(t)=i(t)_{h} +c$

where c is some constant, we added this because the right side of the primary differential equation has a constant term given by V/R. We put this in the main differential equation and obtain the value of c as c=V/R by comparing the constants on both sides.if we put in our initial condition of i(0)=0, we obtain $I_{max} =V/R$ , so the overall equation becomes,

$I(t)=(V/R)(1-e^{-Rt/L})$

where if we just plug in the values given in the question we obtain the answer given below,

$i(t)=(2/3)(1-e^{-300t})$

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3 years ago