Combine the two formulas into a single expression before plugging numbers. The energy is $U=\dfrac{B^{2}}{2\mu_0}V$ with $B=\dfrac{\mu_0 I}{2\pi r}$, so
\[U=\frac{1}{2\mu_0}\left(\frac{\mu_0 I}{2\pi r}\right)^{2}V=\frac{\mu_0 I^{2}}{8\pi^{2} r^{2}}\,V.\]
Now substitute $\mu_0=4\pi\times10^{-7}$, $I=4.00$ A, $r=0.10$ m, $V=10^{-9}\,\text{m}^{3}$:
\[U=\frac{(4\pi\times10^{-7})(16)}{8\pi^{2}(0.01)}(10^{-9}).\]
The numerator is $2.011\times10^{-5}$ and the denominator is $8\pi^{2}(0.01)=0.7896$, giving $2.546\times10^{-5}\,\text{J/m}^{3}$ for the density. Multiplying by $10^{-9}\,\text{m}^{3}$ gives $U\approx2.55\times10^{-14}\,\text{J}$. The tiny volume ($1\,\text{mm}^{3}$) far from the wire makes the stored energy extremely small, confirming option (A).
\[\boxed{U\approx2.55\times10^{-14}\,\text{J}}\]