In the context of stellar bow shock, the magnetopause is the point where the magnetic field is strong enough to brake the incident wind. The condition is that the magnetic pressure at that point is equal to the ram pressure of the wind

$ B_m^2 \approx \rho v^2 $

If the planet can be considered as a magnetic dipole, then

$ B_m \approx B_s \frac{r_s}{r_m} $

where subscript s corresponds to the surface of the planet, and subscript m to the magnetopause. Putting it all together yields

$ r_m \approx r_s \left( \frac{B_s^2}{\rho v^2} \right)^{1/6} $

For example, earth's magnetic field is about $ 45 \mu T $, the proton density of the solar wind is about $ 1.4 cm^{-3} $ and its bulk velocity is about $ 500 \frac{km}{s} $ substituting in the formula above yields

$ r_m \approx 12.5 r_{\oplus} $

where $ r_{\oplus} $ is earth's radius.