Pudding

$a~ is~\bot~base. Let~h~be~in~the~plane~of~a~slant~side~as~a~projection ~of~a,~b~as~a~projection ~of~a~on~the~base.\\ b=1/2*(18-12)=3. ~~~~ a,~ b,~ h, ~form~a~right~\Delta,~~~a, ~and~b~are~the~legs.\\ Due ~to~ equality~of~areas,~~12^2+18^2=4*1/2*(12+18)*h.\\ and~a^2=h^2-b^2.~~~From~the~two~equations, ~we~get,\\$ $a=\sqrt{(\dfrac{12^2+18^2}{4*1/2*(12+18)})^2 - 3^2}=\Large \color{#D61F06}{7.2}.\\ \text{Almost the same as Mr. Chew-Seong Cheong .}$

Let the height of the lateral trapezium be $h$ and the horizontal distance between the top square and bottom square edges be $b$ . Then by Pythagorean theorem we have $a^2+b^2=h^2$ .

Since it is given that the sum of the top and bottom areas equal to the area of the four lateral surfaces, we have $12^2 + 18^2 = 4 \times \dfrac {12+18}2 h$ . $\implies h = 7.8$ . We note that $b = \dfrac {18-12}2 = 3$ , therefore $a= \sqrt{h^2 - b^2} = \sqrt{7.8^2-3^2} = \boxed{7.2}$ .