Transfer function gain
The transfer function shown in the expression you gave must be rewritten to fit a low-entropy format. This format, described by the fast analytical circuits technique or FACTs, will let you gain immediate insight on what the expression does. Here, your original expression is:

\$H(s)=200\frac{s+2}{s(s+20)}\$

Rewrite it by factoring 2 and 20 respectively in the numerator \$N(s)\$ and the denominator \$D(s)\$:

\$H(s)=\frac{200\times 2}{20}\frac{1+\frac{s}{2}}{s(1+\frac{s}{20})}=20\frac{1+\frac{s}{2}}{s(1+\frac{s}{20})}\$ which is the expression given as the answer. This expression can then be rewritten in an advantageous form such as \$H(s)=20\frac{1+\frac{s}{\omega_z}}{s(1+\frac{s}{\omega_p})}\$ in which \$\omega_z=2 \;rad/s\$ which is 318 mHz and \$\omega_p=20\;rad/s\$ which corresponds to \$\approx\$ 3.2 Hz.

The problem I have with this notation is that the leading term, 20, has the dimension of Hz and calling it a gain is improper to me. Furthermore, it has no physical meaning: this transfer function features a pole-zero pair associated with a pole at the origin. The coefficient of \$s\$ in the denominator is \$\frac{1}{20}\$ and has a dimension of time. To build the final transfer function, simply multiply the pole at the origin affected by its coefficient and the pole-zero pair as shown in the below graph:



You see the integrator response which crosses over at 3.2 Hz and the pole-zero pair response which "boosts" the phase between the zero and the pole. What is the significance of the 26-dB "gain" given in the formulated answer? None to my opinion. Actually, look below at the result when combining both responses:



You can see what is called a "type-2" compensator used in control system to boost the phase at the selected crossover frequency, usually where the phase peaks. What matters here is the realization of the so-called mid-band gain occurring around 1 Hz. It is 10 dB, not 26 as given in the correction. How can we unveil it? Simply by using what is called an inverted zero: factor \$\frac{s}{2}\$ in the numerator and simplify the leading term. You have:

\$H(s)=20\frac{1+\frac{s}{2}}{s(1+\frac{s}{20})}=\frac{20}{s}\frac{s}{2}\frac{1+\frac{2}{s}}{1+\frac{s}{20}}=H_0\frac{1+\frac{\omega_z}{s}}{1+\frac{s}{\omega_p}}\$ where, this time, \$H_0=10\$ or 20 dB and has the dimension of a gain. This is the mid-band gain that appears at 1 Hz. It is a design target and you have to design your compensator to a) match the required phase boost at crossover and b) adjust the mid-band gain to the right value. How can you do that easily with the given answer? This answer is mathematically correct but provides 0 insight on the compensator characteristics. Below is the corresponding Mathcad file.