Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
At the instant the switch is closed, the current is assumed to split into two currents, $I_1$ through the left side of the circuit and $I_2$ through the right side of the circuit at which there is an inductor.
But we know that the voltage across the inductor is given by $\Delta V_L=(dI/dt)L$ where at the instant we close the switch $dI/dt$ approaches zero since the change in the current at this instant is infinitesimal.
Note that we built our answer as we assume that the inductor is an ideal one which means that it has no resistance and that the battery has no internal resistance too.
This means that the current at the instant we close works as a wire. So the circuit, at this moment, consists of just the battery and the resistor.
$$\boxed{I_{\rm 0}=\dfrac{\Delta V_{\rm bat }}{R}}$$
$$\color{blue}{\bf [b]}$$
To find $I$ as a function of time $t$, we need to use Kirchhoff’s loop.
Let's start from the lower left corner and move counterclockwise.
$$\Delta V_L+\Delta V_R-\Delta V_{\rm bat}=0$$
$$L\dfrac{dI}{dt}+IR-\Delta V_{\rm bat}=0$$
Hence,
$$ \dfrac{dI}{dt}=\dfrac{ \Delta V_{\rm bat}}{L}-\dfrac{IR}{L}$$
$$ \dfrac{dI}{dt}=\dfrac{R}{L}\left[ \dfrac{ \Delta V_{\rm bat}}{R}-I\right]$$
Plugging from the boxed formula above,
$$ \dfrac{dI}{dt}=\dfrac{R}{L}\left[I_0-I\right]$$
Separating the time and the current:
$$ \dfrac{dI}{\left[I_0-I\right] }=\dfrac{R}{L}dt$$
Integrating, from $t=0$ to $t=t$, and from $I=0$ to $I=I$,
$$ \int_0^I\dfrac{dI}{\left[I_0-I\right] }=\int_0^t\dfrac{R}{L}dt$$
$$ -\ln\left[I_0-I\right]_0^I =\left( \dfrac{R}{L}\right)t $$
$$ \ln\left[\dfrac{I_0-I}{I_0}\right] =-\left( \dfrac{R}{L}\right)t $$
Hence,
$$ \dfrac{I_0-I}{I_0} =e^{-Rt/L} $$
$$ 1- \dfrac{ I}{I_0} =e^{-Rt/L} $$
$$\boxed{ I =I_0\left[1-e^{-Rt/L} \right]}$$
$$\color{blue}{\bf [c]}$$
It is an exponential function, see the graph below.