Answer:
Let the volume of the block of wood be 
V
c
m
3
 and its density be 
d
w
g
c
m
−
3
So the weight of the block 
=
V
d
w
g
 dyne, where g is the acceleration due to gravity 
=
980
c
m
s
−
2
The block floats in liquid of density 
0.8
g
c
m
−
3
 with 
1
4
t
h
 of its volume submerged.So the upward buoyant force acting on the block is the weight of displaced liquid
=
1
4
V
×
0.8
×
g
 dyne.
Hence by cindition of floatation
V
×
d
w
×
g
=
1
4
×
V
×
0.8
×
g
⇒
d
w
=
0.2
g
c
m
-3
,
Now let the density of oil be 
d
o
g
c
m
-3
The block floats in oil with 60% of its volume submerged.So the buoyant force balancing the weight of the block is the weight of displaced oil = 
60
%
×
V
×
d
o
×
g
 dyne
Now applying the condition of floatation we get
60
%
×
V
×
d
o
×
g
=
V
×
d
w
×
g
⇒
60
100
×
V
×
d
o
×
g
=
V
×
0.2
×
g
⇒
d
o
=
0.2
×
10
6
=
1
3
=
0.33
g
c
m
−
3
Step-by-step explanation: