There should be more clearness in the given so that the reader would understand with less confusion. Anyway , this is how I believe the problem should be solved.
At the outset, the bicycle tire has some air in it , the pressure of the gas should be equal to the atmospheric pressure = 1 atm as stated earlier. ( Part of the given, can be computed from the gauge pressure ).
Imagine certain amount of air flows from the atmosphere to fill the tire. Eventually the tire becomes filled with air of gauge pressure = 5 atm. The initial volume of the gas is unknown , V1, but its final volume V2 should be equal to the volume of the tire ( ideal gas ). The gauge pressure is due to the total pressure of the air inside the tire.
I talk about "total" pressure of the gas inside because I assume there is the initial air present ( of partial pressure 1 atm ) and the new air that has passed in of partial pressure P.
Total pressure inside the tire , Pt = G + Patm = 5 atm + 1atm = 6 atm
But the tire already had certain amount of air of partial pressure 1 atm , so the partial of pressure of the air that has been transfered is P = Total pressure inside the tire - partial pressure of air already present = 6 - 1 = 5 atm
The process is isothermal and the number of particles of the moving is constant ( N is fixed ) , so the following holds : P1V1 = P2V2
P1 = 1 atm , the intial pressure of the gas ( atm pressure )
V1 = initial volume , unknown
P2= final partial pressure of the gas transfered = 5 atm
V2 = final volume occupied by the gas = 0.98 L
I hope this helps.