From volume to surface area

Here : $\large \text{Volume of Cube = Volume of cylinder + Volume of Cuboid}$ Now in the cuboid , the third dimension comes out to be 8 inches by applying pythagoras in the diagonal . Hence it is $12\times 2 \times 8$ dimensioned(inches).

Now volume of cuboid = $V_1 = \dfrac{12}{12} \cdot \dfrac{2}{12} \cdot \dfrac{8}{12} = \boxed{\dfrac19 } \text{cube feet}$

Similarly volume of cylinder = $V_2 = \pi r^2 h = \dfrac{22}{7} \cdot \left (\frac{2}{12} \right)^2 \cdot 1 =\boxed{ \dfrac{11}{126}} \text{cube feet}$ Now adding both we get $\dfrac{11}{126} + \dfrac{1}{9} \approx \dfrac15$

Therefore , Volume of cube $\rightarrow$ $V_3 = a^3 = \dfrac15$ $\implies a = \dfrac1{\sqrt[3]{5}}$ Hence required surface area $\rightarrow$ $\implies S = 6a^2 = \dfrac{6}{5^{\frac23}} \approx \color{#EC7300}{\boxed{\text{2 square feet}}}$