Use the hydrogen energy-level formula

For hydrogen atom, $E_n = -\frac{13.6}{n^2}\,\text{eV}$

Given, $E_n = -0.85\,\text{eV}$

So, $-\frac{13.6}{n^2} = -0.85$

Find the principal quantum number $n$

$\frac{13.6}{n^2} = 0.85$ $n^2 = \frac{13.6}{0.85} = 16$ $n = 4$

So the electron is in the $4^{\text{th}}$ excited state? No — it is in the $n=4$ level, which is the third excited state.

Find the maximum number of possible transitions to lower levels

If the atom has electrons in level $n$, then the maximum number of spectral transitions possible among all lower levels from that state is $N = \frac{n(n-1)}{2}$

For $n=4$, $N = \frac{4\cdot 3}{2} = 6$

These transitions are: $4\to 3,\;4\to 2,\;4\to 1,\;3\to 2,\;3\to 1,\;2\to 1$

Final answer

The maximum number of allowed transitions is $\boxed{6}$