Solution:

Let the length, width and height of the rectangular prism made by the 24 cubes without paint be denoted by $\ell, w, h$ (necessarily positive integers), respectively. Then, those of the prism made by the 120 cubes have measures $\ell+2, w+2$ and $h+2$, respectively. Hence, $\ell w h=24$ and $(\ell+2)(w+2)(h+2)=120=2^{3} \cdot 3 \cdot 5$. WLOG, assume that $5$ divides $\ell+2$.

If $\ell+2=5$, then $\ell=3$ and we get $w h=8$ and $(w+2)(h+2)=24$. The values $\{w, h\}=\{4,2\}$ satisfy the problem constraints. In this case, the surface area is
$$2((\ell+2)(w+2)+(w+2)(h+2)+(\ell+2)(h+2))=2(30+20+24)=148.$$

If $\ell+2=10$, then $\ell=8$ and so $w h=3$ and $(w+2)(h+2)=12$. The former implies $\{w, h\}=\{1,3\}$ which does not satisfy the latter constraint.

If $\ell+2 \geq 15$, then $\ell \geq 13$ and therefore, $24 \geq 13 w h$. This forces $w=h=1$ which makes $120=(\ell+2)(w+2)(h+2)$ divisible by $9$, contradiction.