Pressure of argon = 546.8 kPa
Conversion factor: 1 atm = 101.325 kPa
Pressure of argon = 546.8 kPa x 1 atm/101.325 kPa = 5.4 atm
Moles of argon = 15.82
Volume of argon = 75.0 L
According to Ideal gas law,
PV = nRT
where P is the pressure, V is the volume , n is the number of moles, R is the universal gas constant, and T is the temperature
T = PV/nR = (5.4 atm x 75.0 L) / (15.82 x 0.0821 L.atm.mol⁻¹K⁻¹)
T = 311.82 K
Hence the temperature of the canister is 311.82 K.