Using the law of conservation of momentum, total momentum, before and after the impact stays the same. That is
MbVb + MwVw = (Mb + Mw)V
where, Mb and Mw are the masses of the bullet and wood block; Vb and Vw are velocities of bullet and wood block and V is the velocity of combined system (after bullet gets embedded in the block).
The friction force can be calculated as:
F = `mu` (Mb + Mw)g
and the work done by friction force is calculated as:
W = `mu` (Mb + Mw)g x distance
This work done by friction force opposes the motion of bullet + block system (and is balanced by its kinetic energy),
Hence, `1/2` (Mb + Mw) `V^2` = `mu` (Mb + Mw) g x distance
or, `V^2` = 2`mu` g x distance = 2 x 0.3 x 9.81 x 10
or, Initial velocity of the wood + bullet system, V = 7.67 m/s
and using this value in the first equation, we get
0.015 x Vb + 1.25 x 0 = (0.015 + 1.25) x 7.67
thus, Vb = 646.84 m/s
Hence the muzzle velocity of bullet is 646.84 m/s.
Hope this helps.