similarity

I did it like this: 81 and 24, being mass values, were in a cubed ratio, meaning that the 'linear ratio' (so to speak) would be $\sqrt [ 3 ]{ 24 } :\sqrt [ 3 ]{ 81 }$

and the 'squared' ratio (the one relevant to the surface area) would be $\sqrt [ 3 ]{ 24 } ^{ 2 }:\sqrt [ 3 ]{ 81 } ^{ 2 }$

$\sqrt [ 3 ]{ 81 } ^{ 2 }$ turns out to be 18.72. I divided the surface area, 540, by this number, and got 28.845. Now, all that remains is to find the value of

$\sqrt [ 3 ]{ 24 } ^{ 2 }$ which is 8.32, and multiply it by the factor 28.845, which gives us 240.

If two objects are similar, we can say that They have the same form and Theirs measurements are proportionals. Applying the density's formula, $\quad d(x)\quad =\quad \frac { M(x) }{ V(x) }$ , at the objects, we find out the relation $\frac {M(O) }{ M(o)} =\frac {V(O)}{V(o)}$ . This means that the ratio of the volume between the larger and smaller objects is: $\frac { V(O) }{ V(o) } =\frac { 81 }{ 24 } =\frac { { 3 }^{ 4 } }{ { 2 }^{ 3 }\times 3 } ={ \left( \frac { 3 }{ 2 } \right) }^{ 3 }$ .

Having the volumes' ratio, now it's necessary to have the relation between the the volume and surface area. For this, we can use two relations: edge-area and edge-volume, that are, respectively: ${ \left( \frac { { E }_{ n }(O) }{ { E }_{ n }(o) } \right) }^{ 3 }=\frac { V(O) }{ V(o) } ;\quad { \left( \frac { { E }_{ n }(O) }{ { E }_{ n }(o) } \right) }^{ 2 }=\frac { S(O) }{ S(o) }$

For the relation be done, the both equalities must have an commom factor; in this case will be ${ \left( \frac { { E }_{ n }(O) }{ { E }_{ n }(o) } \right) }^{ 6 }$ . For this, the relations will be raised by the, respectively, 2nd and 3th power, and finally equalizing them: ${ \left( \frac { { E }_{ n }(O) }{ { E }_{ n }(o) } \right) }^{ 6 }={ \left( \frac { V(O) }{ V(o) } \right) }^{ 2 }={ \left( \frac { S(O) }{ S(o) } \right) }^{ 3 }$ .

Substituting $\frac { V(O) }{ V(o)}\ )$ by ${ \left( \frac { 3 }{ 2 } \right) }^{ 3 }$ and $S(O)$ ) by ${ 2 }^{ 2 }\times { 3 }^{ 3 }\times 5$ and isolating $S(o)$ , we will have: ${ \left( \frac { 3 }{ 2 } \right) }^{ 6 }={ \left( \frac { { 2 }^{ 2 }\times { 3 }^{ 3 }\times 5 }{ S(o) } \right) }^{ 3 } { \left( \frac { 3 }{ 2 } \right) }^{ 2 }=\frac { { 2 }^{ 2 }\times { 3 }^{ 3 }\times 5 }{ S(o) } \frac { { 3 }^{ 2 } }{ { 2 }^{ 2 } } =\frac { { 2 }^{ 2 }\times { 3 }^{ 3 }\times 5 }{ S(o) } S(o)={ 2 }^{ 4 }\times { 3 }^{ 1 }\times { 5 }^{ 1 }=240$

Thus, The smallest object's surface area is $240㎠$ .

Subtitle adopted: $O$ - Largest object; $o$ - Smallest object; $d(x)$ - Density of an $x$ object; $M(x)$ - Mass of an $x$ object; $V(x)$ - Volume of an $x$ object; ${ E }_{ n }(x)$ -width of n edge of an $x$ object; $S(x)$ - Surface area of an $x$ object

Ratio of linear dimensions is (24/81)^(1/3) = 2/3.
Ratio of surface area is (2/3)^2 = ( ?/540)..........? =240

$\sqrt[3]{[540*540*540/(81*81)]*24*24}$ = 240