Answer
$1.66\times 10^{-35}\;m$
Work Step by Step
The de Broglie wavelength $(\lambda)$ is express by the formula,
$\lambda=\frac{h}{p}$,
where $h$ is the Planck's constant and $p$ is the momentum of the moving particle.
Here, the momentum of the bullet is given by
$p=mv=(0.04\times 1000)\;kg.m/s=40\;kg.m/s$
Therefore, the de Broglie wavelength of the bullet is given by
$\lambda=\frac{6.63\times 10^{-34}}{40}\;m$
or, $\lambda\approx 1.66\times 10^{-35}\;m$
