Help with finding the value of current through a resistor in an RL circuit at a specific point in time
I hope that you can readily follow the schematics and their transitions, below:



simulate this circuit &ndash; Schematic created using CircuitLab
Prior to the switch being opened at \$t=0^+\$, \$L_1\$ has a fixed current in it and therefore has zero voltage across it. I've also then swapped positions of \$V_2\$ and \$R_2\$ (which is perfectly valid for a series pair like this) and so, on the right, you see the equivalent circuit to analyze in order to work out the steady state current in \$L_1\$ at \$t=0^-\$. This is, clearly, just \$750 \:\text{mA}\$, and in the direction shown.
Upon opening the switch, \$V_1\$ and \$R_1\$ are no longer relevant. So they can be removed, entirely. What remains, at \$t=0^+\$, is then shown on the left side of the bottom row. It's equivalent circuit is then shown on the right. (I've chosen a convenient place for ground on the left.) You should be able to see this transformation and agree with it, as well.
This is really a very simple remaining circuit to analyze. The nodal equation is:
$$\frac{V_x}{R_\text{SUM}}+\frac1{L_1}\int V_x\:\text{d}t=\frac{6\:\text{V}}{R_\text{SUM}}$$
Now, we take the derivative of the above in order to help simplify the solution and arrive at:
$$\begin{align*}
\frac{\text{d}}{\text{d}t}\left[\frac{V_x}{R_\text{SUM}}+\frac1{L_1}\int V_x\:\text{d}t\right.&amp;=\left.\frac{6\:\text{V}}{R_\text{SUM}}\right]\\\\
\frac{\text{d}}{\text{d}t}\left[\frac{V_x}{R_\text{SUM}}+\frac1{L_1}\int V_x\:\text{d}t\right]&amp;=\frac{\text{d}}{\text{d}t}\left[\frac{6\:\text{V}}{R_\text{SUM}}\right]\\\\
\frac{1}{R_\text{SUM}}\frac{\text{d}}{\text{d}t}V_x+\frac1{L_1}V_x&amp;= 0\:\text{A}
\end{align*}$$
The solution to the above (integrating factor or whatever else you may prefer) yields:
$$V_x=C_0\cdot e^{\left[-\frac{R_\text{SUM}}{L_1}\cdot t\right]}$$
We know that \$V_x=18\:\text{V}\$ at \$t=0^+\$ (you should easily see why.) Therefore, \$C_0=18\:\text{V}\$. At this point you have everything you need. You are  trying to compute:
$$I\left(t=70\:\text{ms}\right)=\frac{+6\:\text{V}-18\:\text{V}\cdot e^{\left[-\frac{R_\text{SUM}=16\:\Omega}{L_1=2\:\text{H}}\cdot \,t=70\:\text{ms}\right]}}{R_\text{SUM}=16\:\Omega}$$
Which should be very easy to compute.