Answer
$7.5\;\rm cm$
Work Step by Step
As we see below, the center of the mass of the man's arms moves up a distance of 75 cm since the center of the mass of the arm is at its geometric center since the author told us to represent it as a uniform cylinder of mass 3.5 kg.
We need to find the difference between the initial center of the mass of the man's whole body and the final position of his center of mass, as seen in the figure below.
$$\Delta y_{cm,body}= y_{f,cm,body}- y_{i,cm,body}$$
where $i$ is when the arms are down and $f$ is when the arms are up.
We can call the rest of the body, without the two arms, the NoArm body.
$$\Delta y_{cm,body}= \dfrac{(m)_{NoArm} (y_{f,cm})_{NoArm }+2m_{arm}y_{f,arm}}{(m)_{NoArm}+2(m)_{arm}} \\
- \dfrac{(m)_{NoArm} (y_{i,cm})_{NoArm }+m_{arm}y_{i,arm}}{(m)_{NoArm}+2(m)_{arm}}
$$
Noting that $[(m)_{NoArm}+2(m)_{arm}]$ is the mass of the whole body $M$. Also, note that $2(m)_{arm}$ is plugged since we have two arms, not one.
$$\Delta y_{cm,body}= \dfrac{(m)_{NoArm} (y_{f,cm})_{NoArm }+2m_{arm}y_{f,arm}}{M} \\
- \dfrac{(m)_{NoArm} (y_{i,cm})_{NoArm }+2m_{arm}y_{i,arm}}{M}
$$
$$\Delta y_{cm,body}= \dfrac{ 1}{M}\left[\overbrace{ (m)_{NoArm} (y_{f,cm})_{NoArm }
-(m)_{NoArm} (y_{i,cm})_{NoArm } }^{=0} \\
+ 2m_{arm}y_{f,arm} \\
- 2m_{arm}y_{i,arm} \right]
$$
We can see that the center of the mass of the NoArm part of the body is stationary since nothing happened to it.
Thus,
$$\Delta y_{cm,body}= \dfrac{ 1}{M}\left[ 2m_{arm}y_{f,arm} - 2m_{arm}y_{i,arm} \right] $$
$$\Delta y_{cm,body}= \dfrac{ 2m_{arm}}{M}\left[ y_{f,arm} - y_{i,arm} \right] $$
From the figure below, we can see that $ y_{f,arm} - y_{i,arm} =2(\frac{1}{2}L_{arm})=75$ cm.
Hence,
$$\Delta y_{cm,body}= \dfrac{ 2m_{arm}}{M}\left[ 0.75\right] $$
Plugging the known;
$$\Delta y_{cm,body}= \dfrac{ 2\times 3.5}{70}\times \left[ 0.75\right] =\bf 0.075\;\rm m$$
$$\Delta y_{cm,body}= \color{red}{\bf 7.5}\;\rm cm$$
