Electrical – Voltage across inductor immediately after the switch is closed (again)

So the 3, 6 and 12 ohms resistors are in parallel (asuming the inductor is 0 ohms) - this means \${12 \over 7} ohms\$. Add the 4 ohms resistor and you got \$ {12 \over 7} + 4 = {40 \over 7} ohms\$. The current through this is \$60 \cdot {7 \over 40} = 10.5 A\$. Using this we calculate the voltage across the three resistors in parallel:

\$ {12 \over 7} \cdot 10.5 = 18 V\$. Then divide this by the 3 ohms resistor: \${18 V \over 3 ohms} = 6 A\$.

^{simulate this circuit – Schematic created using CircuitLab}

The correct answers were already given. As a supplement, I will try to give a formal mathematical derivation.

To begin with, you can integrate both sides of the equation to get a formula for the current through an inductor at time \$t\$:

$$V(t) = L\frac{\mathrm{d}I(t)}{\mathrm{d}t} \implies \int_0^t V(\tau)\,\mathrm{d}\tau = LI(t) \implies I(t) = \frac{1}{L}\int_0^t V(\tau)\,\mathrm{d}\tau$$ here we assume that \$I(0) = 0\$, otherwise add \$I(0)\$ to the expression for \$I(t)\$.

"The average voltage across an ideal inductor is always zero" actually means the average voltage over a period is zero (otherwise it's meaningless to impose such condition). That is, here we assume that the voltage across an inductor is periodic.

Assume that the voltage across an inductor is a periodic function with period \$T\$. "Is periodic with period \$T\$" is just another way to say that \$V(t) = V(t + T)\$ for any \$t\$.

Let's calculate the current after \$n\$ periods, i.e. when \$t = nT\$ (\$n\$ is an integer):

$$ I(nT) = I(\underbrace{T + T + \dots + T}_{\text{n times}}) = \frac{1}{L}\int_0^{\underbrace{T + T + \dots + T}_{\text{n times}}} V(\tau)\,\mathrm{d}\tau$$

Here we can use the property of an integral \$\int_0^{x+y} f(t)\,\mathrm{d}t = \int_0^x f(t)\,\mathrm{d}t + \int_x^y f(t)\,\mathrm{d}t\$ to break the integral into a sum:

$$ \begin{split} I(nT) &= \frac{1}{L}\left( \underbrace{ \int_0^T V(\tau)\,\mathrm{d}\tau + \int_T^{2T} V(\tau)\,\mathrm{d}\tau + \dots + \int_{(n-1)T}^{nT} V(\tau)\,\mathrm{d}\tau }_{\text{n times}} \right)\\ &= n\cdot\frac{1}{L}\int_0^T V(\tau)\,\mathrm{d}\tau \quad \text{because }V(\tau)\text{ is periodic with period }T\\ &= n\cdot I(T) \end{split} $$

From this expression you can see that if the integral over a whole period $$I(T) = \frac{1}{L}\int_0^T V(\tau)\,\mathrm{d}\tau$$ is not zero, then after \$n\$ periods the current through an inductor will be n times larger: $$\boxed{I(nT) = n\cdot I(T)}$$ As \$n\$ goes to infinity, so does the the current.

Thus, the only way to keep current from going to infinity is the condition \$I(T) = 0\$, which is equivalent to $$\int_0^T V(\tau)\,\mathrm{d}\tau = 0$$ because \$\frac{1}{L}\$ is just a constant factor. Just to remind you, \$T\$ is the period of the voltage. The lower limit of integration is \$t = 0\$. "Zero time" can be an arbitrary chosen instant of time, because the process is periodic.