How dBuV value changes with distance?
The electric field amplitude of an electromagnetic wave falls as 1/d in the far field. If you look at the power density it falls as \$1/d^2\$

Going to dB scale you must do 10log p (p is power density) so it depends on 20log d (because d was squared). The dB increase is therefore 20log(3/1)=9.5 dB.

Note that the previous calculation is only valid if we're working at the far field (boundary between near and far field depends on \$\lambda\$, i.e. on the transmitter frequency). In the near field you'll have inductive or capacitive coupling between transmitter and receiver, and not simply radiative coupling, so that the power increase by approaching the antennas would be much bigger.