A solid cube is cut into two cuboids of equal volumes. The ratio of the total surface area of the given cube and that of one of the cuboids is

The correct option is A: $6a^2:4a^2$

Let the edge of the solid cube $=a$
Therefore volume of the cube $=a^3$

Sine, it is cut into cuboids of equal volume, therefore, volume of each cuboid is $\frac{a^3}{2}$

This is possible when Length $=$ Height $=a$ and breadth $=\frac{a}{2}$

Now, TSA (Total surface area) of the cube $=6\times {\text{side}}^2=6a^2$ and

TSA of a cuboid $=2(lb+bh+lh)$

$=2(a\times \frac{a}{2}+a\times \frac{a}{2}+a\times a)$

$=2(a^2+a^2)$

$=4a^2$

Therefore, the required ratio is $=\frac{6a^2}{4a^2}=3:2$