Regions of electron density can be bonds - simple, double, or triple bonds count as 1 region - or lone pairs of electrons.
Since the steric number determines orbital hybridization, you can work backwards and use orbital hybridization to determine the steric number.
In your case,
That being said, you could eyeball the given molecules and predict which could match this description, even without having to draw their respective Lewis structure.
#SeF_6#, selenium hexafluoride
The selenium atom is bonded to 6 fluorine atoms, and since selenium has 6 valence electrons (it's part of the oxygen family), no lone pairs could possibly be present on it.
As a result, you get 6 bonds and 0 lone pairs
#XeCl_4#, xenon tetrachloride
Xenon is a noble gas, so it has 8 valence electrons. Since it only forms 4 single bonds with the chlorine atoms, which account for 4 of the 8 valence electrons, it will also have 2 lone pairs present.
As a result, you get 4 bonds and 2 lone pairs
#IF_5#, iodine pentrafluoride
Iodine is a halogen, which means it has 7 valence electrons. Since it forms 5 single bonds with the fluorine atoms, it will have 2 valence electrons present as a lone pair.
As a result, you get 5 bonds and 1 lone pair
#AsCl_5#, arsenic pentachloride
Arsenic is part of the nitrogen family, which means it has 5 valence electrons. Since it forms 5 single bonds with the chlorine atoms, you won't have any lone pairs present.
As a result, you get 5 bonds and 0 lone pairs
SIDE NOTE As practice, you could actually draw the Lewis structures for all the molecules and double-check the answer.