As per de Broglie’s equation,

$
\lambda=\frac{h}{m v}
$
Where, $\lambda$ denotes thr wavelength of the moving particle
$\mathrm{m}$ is the mass of the particle
$v$ denotes the velocity of the particle
$\mathrm{h}$ is Planck’s constant
Substituting these values in the expression for $\lambda$ :
$
\begin{array}{l}
\lambda=\frac{\left(6.626 \times 10^{-34}\right) J_{s}}{\left(9.10939 \times 10^{-31} \mathrm{~kg}\right)\left(2.05 \times 10^{7} \mathrm{~ms}^{-1}\right)} \\
=3.548 \times 10^{-11} \mathrm{~m}
\end{array}
$
Therefore, the wavelength associated with the electron which is moving with a velocity of $2.05 \times$
$
10^{7} \mathrm{~ms}^{-1} \text { is } 3.548 \times 10^{-11} \mathrm{~m}
$