The difference between the incident energy and threshold energy for an electron in a photoelectric effect experiment is 5 eV. The de Broglie wavelength of the electron is

The excess of incident energy over threshold energy becomes the maximum kinetic energy of the emitted electron, so $K = 5\,\mathrm{eV}$. Using de Broglie relation, $$\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$$ For an electron, substituting $h = 6.6 \times 10^{-34}\,\mathrm{J\,s}$, $m = 9.1 \times 10^{-31}\,\mathrm{kg}$, and $K = 5 \times 1.6 \times 10^{-19}\,\mathrm{J}$, $$\lambda = \frac{6.6 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 5 \times 1.6 \times 10^{-19}}}$$ Collecting powers of $10$, $$\lambda = \frac{6.6 \times 10^{-34}}{\sqrt{145.6 \times 10^{-50}}}$$ Therefore, $$\lambda = \frac{6.6 \times 10^{-9}}{\sqrt{145.6}}\,\mathrm{m}$$ Comparing with the given options, this matches option $B$.