1. Given data

• Inductance: $L = 2\,\text{H}$
• Applied voltage: $V(t) = 3t\,\text{V}$
• Time: $t = 4\,\text{s}$
• Resistance is negligible, so for an ideal inductor: $V = L\frac{di}{dt}$

2. Find current as a function of time

Using $L\frac{di}{dt} = 3t$ Substitute $L=2$: $2\frac{di}{dt} = 3t$ $\frac{di}{dt} = \frac{3t}{2}$

Integrate with respect to $t$: $i(t) = \int \frac{3t}{2}\,dt = \frac{3}{4}t^2 + C$

Since the voltage is applied at $t=0$, initial current in the inductor is taken as zero: $i(0)=0 \Rightarrow C=0$

So, $i(t)=\frac{3}{4}t^2$

At $t=4\,\text{s}$, $i(4)=\frac{3}{4}(4)^2=\frac{3}{4}\cdot 16=12\,\text{A}$

3. Energy stored in the inductor

Energy stored is $U = \frac{1}{2}Li^2$

Substitute $L=2$ H and $i=12$ A: $U = \frac{1}{2}\cdot 2 \cdot (12)^2$ $U = 144\,\text{J}$

4. Final answer

$\boxed{144}$