Photoelectrons are liberated by ultraviolet light of wavelength 3000Å from a metallic surface for which the photoelectric threshold is 4000Å. The de-Broglie wavelength of electrons emitted with maximum kinetic energy is:

Kinetic energy= Quantum energy - Threshold energy
K.E=6.625×10⁻³⁴×3×10⁸/3000×10⁻¹⁰ - 6.625×10⁻³⁴×3×10⁸/4000×10⁻¹⁰=1.6565×10⁻¹⁹J
1/2mv²=1.6565×10⁻¹⁹
m²×v²=2×1.6565×10⁻¹⁹×9.1×10⁻³¹
mv=5.49×10⁶⁵
λ=h/mv
λ=6.625×10⁻³⁴/5.49×10⁶⁵=1.2×10⁻⁹m
I.e 1.2 nm (1nm=10⁻⁹m)
Hence option A is correct.