SOLUTION: A cylindrical tin 8cm in diameter contains water to a depth of 4cm. If a cylindrical wooden rod 4cm in diameter and 6cm long is placed in the tin, it floats exactly half submerged.

The volume of the water in the tin is (= was originally) 

     =  =  = 201.0619 cm^3  (rounded).


The volume of the half of the cylindrical wooden rod is

     =  = 37.6991 cm^3  (rounded).


The combined volume occupied by the water in the tin and the half of the cylindrical wooden rod 
submerged in the water is this sum 

    201.0619 + 37.6991 = 238.761 cm^3  (rounded).


    +---------------------------------------------------------------------+
    |   Notice that from the numbers given in the problem, it is clear    |
    |   that the submerged rod does not touch the bottom of the tin.      |
    +---------------------------------------------------------------------+


Let x be the level of the water in the tin after submerging half of the cylindrical 
wooden rod (= the final level).


Then we have this equation to find x

     = 238.761 cm^3,  

or

     = 238.761.


It gives

    x =  = 4.75 cm  (rounded).


ANSWER.  The new depth of the water in the tin is  4.75 cm.