Answer
$$0.151 $$
Work Step by Step
From the expression for $n(K)$ given we may write $n(K) \propto K^{1 / 2} e^{-K / k T}$. Thus, with
$$k=8.62 \times 10^{-5} \mathrm{eV} / \mathrm{K}=8.62 \times 10^{-8} \mathrm{keV} / \mathrm{K}$$
We have
$$ \frac{n(K)}{n(K)} =\left(\frac{K}{K_{\mathrm{arg}}}\right)^{1 / 2} e^{-\left(K-K_{\mathrm{ag}}\right) / k T}$$$$=\left(\frac{5.00 \mathrm{keV}}{1.94 \mathrm{keV}}\right)^{1 / 2} \exp \left(-\frac{5.00 \mathrm{keV}-1.94 \mathrm{keV}}{\left(8.62 \times 10^{-8} \mathrm{keV}\right)\left(1.50 \times 10^{7} \mathrm{K}\right)}\right) $$$$=0.151 $$
