How to calculate output resistance in collector circuit?
The output resistance \$r_a\$ is the parallel combination of \$R_E\$ and the resistance looking into the emitter of the transistor. As shown in the solution drawing, the resistance looking into the emitter of the transistor is \$1/g_m\$ plus some other resistance due to the source resistance \$R_G\$ and the biasing resistors \$R_B\$ and \$R_V\$. That "other resistance" caused by \$R_G\$, \$R_B\$, and \$R_V\$ is denoted by a convenience term that the solution calls \$r_a'\$. In equation form, this means that $$ r_a = R_E || \left(\frac{1}{g_m}+r_a'\right) $$

Now we need to calculate \$r_a'\$. In small signal analysis the DC sources are turned off so \$U_0 = U_G = 0\$, and this means that \$R_G\$, \$R_B\$, and \$R_V\$ end up in parallel with each other. The parallel combination of these three resistors is shown in the solution in the dotted box with the value \$R\$. \$r_a'\$ is the parallel combination of these resistors but divided by a factor of approximately \$\beta\$, the small signal current gain (since the current through the three resistors is \$i_b\$ but we were calculating the resistance looking into the emitter). Therefore $$ r_a'= \frac{R_G || R_B || R_V}{\beta}$$

You can substitute the equation for \$r_a'\$ into the equation for \$r_a\$ to get $$ r_a = R_E || \left(\frac{1}{g_m}+\frac{R_G || R_B || R_V}{\beta}\right) $$

The transistor model this solution uses is a little different than the commonly used hybrid-\$\pi\$ model which uses a resistance from base to emitter \$r_{\pi} = \beta/g_m\$. You might want to calculate the solution using the hybrid-\$\pi\$ model since this circuit's output resistance is a little easier to express in terms of \$r_{\pi}\$.