Surface area

$\text{surface area of the solid= surface area of the cube + lateral area of the 6 pyramids - base area of the six pyramids}$

$x=\sqrt{4^2-2^2}=2\sqrt{3}$

$L=\sqrt{3^2+(2\sqrt{3})^2}=\sqrt{21}$

base area of the six pyramids: $A_B=6(6)\left(\dfrac{\sqrt{3}}{4}\right)(4^2)=144\sqrt{3}$

lateral area of the six pyramids: $A_L=\dfrac{1}{2}(P)(L)=\dfrac{1}{2}(6)(4)(\sqrt{21})(6)=72\sqrt{21}$

surface area of the cube: $A_S=6(10)^2=600$

$surface~area~of~the~solid=600+72\sqrt{21}-144\sqrt{3}=$ $\boxed{680.5301}$

comments:

1. In order to compute for the slant height (L), I computed first for x.

2. In the computation for the base area of the six pyramids, I computed the area of one regular hexagon then multiplied it by 6.

3. In the computation for the lateral area of the six pyramids, I computed the lateral area of one pyramid then multiplied it by 6. I used the formula $A=\dfrac{1}{2}PL$ where $P$ is the perimeter of the base and $L$ is the slant height. The slant height is the altitude of one face.