A bullet with a mass of 6.00 g is fired through a 1.25 kg block of
wood on a frictionless surface. The initial speed of the bullet is 896
m/s, and the speed of the bullet after it exits the block is 435 m/s.
At what speed does the block move after the bullet passes through it?

Seems like a question from school assignment, so just a hint.
Apply conservation of linear momentum.
Initial momentum of bullet = sum of final momentums of block and bullet

First step convert masses to equivalents.

i.e. 6g bullet becomes 0.06kg

Now momentum p=mv where p = momentum m = mass v = velocity

There will be conservation of momentum pre and post interaction.

Therefore

p(bullet before interaction) = 0.06*896
                             = 53.76

So
53.76 = p(bullet post interaction) + p (block post interaction)
53.76 = (0.06*435) + 1.25v
v     = 22.128 m/s
v     = 22.1 m/s to 3 s.f.