If the kinetic energy of the particle is increased to 16 times its...

$K{E}_{final} = 16K{E}_{initial}$

$⇒ \dfrac{1}{2}m{{v}_{f}}^{2} = 16 \times \dfrac{1}{2}m{{v}_{i}}^{2}$(mass remains same because the particle remains same)

$⇒ {v}_{f} = 4{v}_{i}$
We know that, De-broglie wavelength $( \lambda ) = \dfrac{h}{mv}$

$∴ {\lambda}_{f} = \dfrac{h}{m{v}_{f}} = \dfrac{h}{4m{v}_{i}}$

So De-Broglie wave will decreased to $75$