The expression for the circular path in a uniform magnetic
fields is,

$R=\frac{mv}{qB}$

$mv=qRB$

We know that de-Broglie wavelength of a charged particle is given by,

$\lambda =\frac{h}{mv}$

Substitute the expression in the above equation.

$\lambda =\frac{h}{R{q}_{\alpha }B}$
$\lambda =\frac{h}{(+2e)BR}$
$\lambda =\frac{6.6\times {10}^{-34}}{2\times 1.6\times {10}^{-19}\times 0.125\times 1\times {10}^{-2}}=1.65\times {10}^{-12} \mathrm{m}$