Answer
$1.69\times10^8\;Pa$.
Work Step by Step
Find the number of moles of argon in 105.0 kg.
$$n=\frac{105.0\;kg}{39.95\times10^{-3}kg/mol}=2628\;mol$$
Use the ideal gas model for argon to calculate the pressure of the gas.
$$P=\frac{nRT}{V}=\frac{(2628\;mol)(8.314)(273.15+21.6\;K)}{38.0\times 10^{-3}\;m^3 }\approx 1.69\times10^8\;Pa $$