#### Answer

The proton will experience the same magnitude of force at a distance of 2 cm from the wire.

#### Work Step by Step

The force on the electron is $F = E~q$
The strength of the electric field at a distance $r_1$ from a charged wire is $E_1 = \frac{\lambda_1}{2\pi ~\epsilon_0~r_1}$
Note that an electron and a proton have the same magnitude of charge. After the charge density is doubled, if a proton is to experience the same magnitude of force, the strength of the electric field $E_2$ must be the same as $E_1$.
$E_2 = E_1$
$\frac{\lambda_2}{2\pi ~\epsilon_0~r_2} = \frac{\lambda_1}{2\pi ~\epsilon_0~r_1}$
$\frac{2~\lambda_1}{2\pi ~\epsilon_0~r_2} = \frac{\lambda_1}{2\pi ~\epsilon_0~r_1}$
$r_2 = 2~r_1$
$r_2 = (2)(1~cm)$
$r_2 = 2~cm$
The proton will experience the same magnitude of force at a distance of 2 cm from the wire.