This isn't rocket science, it's rocket math!

Relevant wiki: Surface Area - Problem Solving

NOTE: The figure above is not drawn to scale.

Let $A$ be the surface area of the top part, $B$ be the surface area of the middle part and $C$ be the surface area of the bottom part. Then,

$A=\dfrac{1}{2}cL=\dfrac{1}{2}(\pi)(2)\sqrt{5^2+1^2}=\pi \sqrt{26}$

$B=ch=2\pi(10)=20 \pi$

$C=\dfrac{1}{2}(c_1+c_2)L + \pi r^2=\dfrac{1}{2} \pi(2+4)\sqrt{5^2+1^2} + \pi (2^2)=\dfrac{1}{2}\pi(6)\sqrt{26}+4\pi=3\pi\sqrt{26}+4\pi$

Finally,

$SA=A+B+C=\pi\sqrt{26}+20\pi+3\pi\sqrt{26}+4\pi \approx$ $\color{#D61F06}\boxed{139}$ $\boxed{answer}$

Definitions:

$c$ = circumference of the circle

$L$ = slant height

$r$ = radius of the circle

$h$ = height

$c_1$ = upper circumference

$c_2$ = lower circumference

CSA of the cone portion $=\pi r l = \pi \cdot 1 \cdot \sqrt{h^2+r^2}=\pi \sqrt{26}$

CSA of cylinder portion $=2\pi r h = 2\pi \cdot 1 \cdot 10=20\pi$

CSA of frustum portion $=\pi (r_{1}+r_{2})\sqrt{h^2+(r_{1}-r_{2})^2}+\pi r_{2}^2=3\pi \sqrt{26}+4\pi$

The Total SA, $=4\pi \sqrt{26}+24\pi\sim 139.47 \\ \Rightarrow \lfloor 139.47 \rfloor = \boxed{139}$

The two conical bits (frustum at the bottom and cone at the top) have the same height, and the radius of the top cone is half the radius of the bottom of the frustum. Proportions therefore allow us to put these two pieces together to produce a cone of height 2*5 = 10. The slant length, $l$ , of this cone is $\sqrt{10^{2}+2^{2}}$ , by Pythagoras' Theorem. Since the curved surface area of a cone is $\pi rl$ , we find that it is $2\pi\sqrt{10^{2}+2^{2}}$ .

Next, the area of a circle is $\pi r^{2}$ , so the area of the base is $4\pi$ .

Finally the curved surface area of a cylinder is $\pi r^{2}h$ . This is $20\pi$ here.

The total is $2\pi \sqrt{104} + 4\pi + 20\pi$ .

This value is just short of $140$ , hence the answer of $139$ .

you should ve wrote that you must round it