Answer
$2.2\times 10^{-3}M/atm$
Work Step by Step
We know that
$n_{Ar}=\frac{P_{Ar}V_{Ar}}{RT}$
$\implies n_{Ar}=\frac{1.0atm\space mol.K(0.0537L)}{(0.08206L.atm)(298.15K)}=2.1949\times 10^{-3}mol$
Now $S_{Ar}=\frac{n_{Ar}}{V_{soln}}$
$S_{Ar}=\frac{2.1949\times 10^{-3}mol}{1.0L}=2.1949\times 10^{-3}M$
We can calculate the Henry's law constant as
$k_H=\frac{S_{Ar}}{P_{Ar}}$
We plug in the known values to obtain:
$k_H=\frac{2.1949\times 10^{-3}M}{1.0atm}=2.2\times 10^{-3}M/atm$
