Answer
$\lambda = 3.313\times 10^{-27}~m$
A proton is larger than this de Broglie wavelength by a factor of $3.02\times 10^{11}$
Work Step by Step
We can find the de Broglie wavelength:
$\lambda = \frac{h}{mv}$
$\lambda = \frac{6.626\times 10^{-34}~J~s}{(1.0\times 10^{-4}~kg)(2\times 10^{-3}~m/s)}$
$\lambda = 3.313\times 10^{-27}~m$
We can find the ratio of a proton to this de Broglie wavelength:
$\frac{1\times 10^{-15}~m}{3.313\times 10^{-27}~m} = 3.02\times 10^{11}$
A proton is larger than this de Broglie wavelength by a factor of $3.02\times 10^{11}$.
