K limiting disturbance effect to 2%

With $R = 0$, the output due to disturbance is found from the loop with $G_2$ forward and $G_1$ in feedback; then the final value theorem gives its steady state.

Let $G_1 = \dfrac{5K}{s+10}$ and $G_2 = \dfrac{6}{s+2}$.

Transfer function from $D$ to $C$ (set $R = 0$)

$$\frac{C(s)}{D(s)} = \frac{G_2}{1 + G_1G_2} = \frac{\frac{6}{s+2}}{1 + \frac{30K}{(s+10)(s+2)}} = \frac{6(s+10)}{(s+10)(s+2) + 30K}$$

Steady state for a step disturbance $D(s) = d/s$

$$c_{ss} = \lim_{s\to 0} s\cdot\frac{d}{s}\cdot\frac{6(s+10)}{(s+10)(s+2)+30K} = \frac{60\,d}{20 + 30K} = \frac{3d}{1 + 1.5K}$$

Condition: $c_{ss} \le 2\%$ of $d$

$$\begin{aligned} \frac{3}{1 + 1.5K} &\le 0.02 \\ 1 + 1.5K &\ge 150 \\ K &\ge \frac{149}{1.5} = 99.33 \end{aligned}$$

Stability check: characteristic equation $s^2 + 12s + 20 + 30K = 0$; all coefficients positive for $K > 0$, so the second order system is stable for this $K$.

Answer: $K \ge 99.33$ (about $K = 100$) keeps the steady state output due to the disturbance within 2% of $d(t)$.