One Digital Logic Question in Previous Exam?
If \$c=0\$, then the top tri-state is hi-Z. The lower tri-state must be on (\$a=1\$) for the output to be well-defined, so \$out=c=0\$.

If \$c=1\$,  then the top tri-state is enabled and:

a. \$a = 0\$ so that the bottom tri-state is off and \$out=a\cdot b = 0\$; or
  b. \$a = 1\$ and the top and bottom tri-state must have the same output for the state to be valid, that is to say, \$\overline{c} = \overline{a\cdot b}\$. Given the known values, \$b = 1\$ is required and \$out = 1\$

This analyses all possible valid combinations (based on the valid possible configurations of the tri-state buffers). We can see that the output is 1 when \$a\cdot c\$, and 0 otherwise. Thus,

\$out = a \cdot c\$

This is perhaps not the most efficient way of analysing it, or the most generalisable (truth tables and then a Karnaugh map reduction would be more general), but it seems a better approach to understanding what's going on intuitively.

Point b above is potentially dangerous in actual implementation, due to shoot-through in the case that \$a = c = 1\$ and \$b = 0\$.