Answer
$\lambda = 6.2 \mathrm{fm}$
Work Step by Step
The momentum times the speed of the light is
$$p c=\sqrt{K^{2}+2 K m c^{2}}=\sqrt{(200 \mathrm{MeV})^{2}+2(200 \mathrm{MeV})(0.511 \mathrm{MeV})}=200.5 \mathrm{MeV}$$
Hence, the wavelength is
$$\lambda=\frac{h c}{p c}=\frac{1240 \mathrm{eV} \cdot \mathrm{nm}}{200.5 \times 10^{6} \mathrm{eV}}=6.18 \times 10^{-6} \mathrm{nm} \approx \boxed{6.2 \mathrm{fm}}$$
