I find this problem numerically inconsistent.
given that P(odd)=90% of P(even)
=> P(even)+0.9*P(even)=1
=> P(even)=0.5263
Also, since all even faces are equally likely,
So, P(2)=P(4)=P(6)=0.5263/3=0.1754 --(1)
Now,
P(the face is even given that the face is greater than 3)=0.75
=> P(the face is 4 or 6)=0.75 => P(4)+P(6)=0.75
from (1)
0.1754+0.1754=0.75
=> 0.3508=0.75
Which means the data is wrong.