An electron in a hydrogen atom has a total energy of -0.85 eV ....

The total energy of an electron in the $n^{\text{th}}$ orbit of a hydrogen atom is given by $E = \dfrac{-13.6}{n^2} \text{ eV}$.

Given $E = -0.85 \text{ eV}$, we have:
$\dfrac{-13.6}{n^2} = -0.85$
$n^2 = \dfrac{13.6}{0.85} = 16$
$n = 4$

According to Bohr's postulate, the angular momentum of the electron is $L = \dfrac{nh}{2\pi}$.

Substituting $n = 4$, we get:
$L = \dfrac{4h}{2\pi} = \dfrac{2h}{\pi}$

Answer: $\dfrac{2h}{\pi}$