The equation that allows you to calculate a particle's DeBroglie wavelength looks like this
#lamda = h/p#, where
#lamda# - the wavelength of the particle, in your case of the electron;
#h# - Planck's constant, equal to #6.63 * 10^(-34)"J s"#;
#p# - the momentum of the electron.
The momentum of the electron is defined as the product between its mass, which is equal to #9.1 * 10^(-31)"kg"#, and velocity, which you know to be 1% of the speed of light.
The velocity of the electron will thus be
#v = 1/100 * c = 1/100 * 3 * 10^(8)"m/s" = 3 * 10^(6)"m/s"#
The momentum of the electron will be
#p = m * v#
#p = 9.1 * 10^(-31)"kg" * 3 * 10^(6)"m"/"s" = 2.73 * 10^(-24)("kg m")/"s"#
This means that the wavelength of the electron will be
#lamda = h/v = (6.63 * 10^(-34)overbrace("J")^(color(blue)("kg m"^2/"s"^2)) "s")/(2.73 * 10^(-24)"kg m"/"s"#
#lamda = 2.43 * 10^(-10) (cancel("kg") "m"^(cancel(2))/cancel("s"^2))/((cancel("kg") cancel("m"))/(cancel("s"))) = color(green)(2.43 * 10^(-10)"m"#
