Step 1: Formula
Use $U = \dfrac{LI^2}{2}$.
Step 2: Square the current
$I^2 = 2.25\ \text{A}^2$.
Step 3: Evaluate
$U = \dfrac{4\times2.25}{2} = 4.5$ J.
Step 4: Convert
$1\ \text{J} = 1000\ \text{mJ}$, so $U = 4500$ mJ.
Step 5: Unit conversion and common slips
Energy in an inductor is easily misread by a factor of 1000. Here $L$ is in henry and $I$ is in ampere, so $\frac{1}{2}LI^2$ comes out directly in joule. Only after getting 4.5 J do we convert to millijoule by multiplying by 1000. Option (A), 4.5 mJ, and option (C), 45 microjoule, are both far too small for a 4 H inductor carrying 1.5 A.
Final Answer:
The energy is 4500 mJ. This is option (D).
\[ \boxed{\text{(D) }4500\ \text{mJ}} \]