Its Not that easy!!

For the contact force we will have to calculate the resultant of the Frictional force as well as the Normal force.

The Frictional Force would be = $\mu mg \cos \theta$

= $0.5 \times 2 \times 10 \times \frac {\sqrt3}{2}$

= $5\sqrt3$

The Normal force = $mg \cos \theta$

= $2 \times 10 \times \frac {\sqrt3}{2}$

= $10\sqrt3$

The Resultant = $\sqrt { { \left( 5\sqrt { 3 } \right) }^{ 2 }+{ \left( 10\sqrt { 3 } \right) }^{ 2 } }$

= $\sqrt {375}$

Hence, answer = $\boxed {\huge 15}$

Let Us Imagine A Block Sliding Down The Plane. Be Careful Here! It is saying "THE MAGNITUDE OF CONTACT FORCE" between the plane and block.Contact force means the force provided by the plane to the block.If We Analyze This We will see that two forces are being provided to the block by the plane.One is the Frictional Force And Other the normal reaction.so contact force would be the Resultant of frictional force and the normal reaction.(the angle between them is 90 degrees.Resolving Forces Parallel and perpendicular to the plane. N = mgcos30 and Frictional force = 0.5*N.Taking Resultant Using laws of vector algebra and solving We Get $\sqrt{375}$ . So a = 375 and sum of digits = 3+7+5 = 15 Ans.