Answer
$coefficient\, of\, friction = tan\theta$
Work Step by Step
The weight can be divided into two components one parallel to the inclined snow surface and the other perpendicular to it.
The component parallel to the surface will be equal to frictional force.
And the other component perpendicular to the surface will be equal to the reaction force.
So, if $\theta$ is the inclination,
$mgsin\theta=coefficient\, of\, friction \times N$
or, $mgsin\theta=coefficient\, of\, friction \times mgcos\theta$
or, $coefficient\, of\, friction = tan\theta$
