The given angular momentum of the electron is $$1.5\left(\frac{h}{\pi}\right)$$, where $h$ is Planck's constant.

According to the Bohr model of the hydrogen atom, the allowed angular momenta for an electron are quantized and given by $$mvr = n \frac{h}{2\pi}$$ where $m$ is the mass of the electron, $v$ is its velocity, $r$ is the radius of the orbit, and $n$ is the principal quantum number (an integer).

By comparing the given angular momentum with the quantized form, we get $$1.5 \frac{h}{\pi} = n \frac{h}{2\pi}$$, which simplifies to $n = 3$.

Now, we use the formula for the energy levels of the hydrogen atom: $$E_n = -\frac{13.6 \text{eV}}{n^2}$$. Since $n = 3$, the energy can be calculated as $$E_3 = -\frac{13.6 \text{eV}}{3^2}$$, which is approximately $-1.5 \text{eV}$.