Answer
*When driven frequency is $10\%$ above $\omega_{0}$ amplitude $=\frac{2A}{3}$
*When driven frequency is $10\%$ below $\omega_{0}$ amplitude $=\frac{A}{1.3}$
Work Step by Step
Let's assume $A(\omega_{d})=A$, when $\omega_{d}=\omega_{0}$. So we can write,
$A=\frac{F_{d}}{m\sqrt {(\omega_{d}^{2}-\omega_{0}^{2})^{2}+\omega_{0}^{2}\omega_{0}^{2}/25}}=\frac{5F_{d}}{m\omega_{0}^{2}}$
When driven frequency is $10\%$ above $\omega_{0}$
$A(\omega_{d1})=\frac{F_{d}}{m\sqrt {(1.1^{2}\omega_{0}^{2}-\omega_{0}^{2})^{2}+\omega_{0}^{2}\times1.1^{2}\omega_{0}^{2}/25}}$
$A(\omega_{d1})=\frac{F_{d}}{m\sqrt {0.04\omega_{0}^{4}+\omega_{0}^{4}\times(\frac{1.1}{5})^{2}}}= \frac{5F_{d}}{m\omega_{0}^{2}\sqrt {1+1.21}}$
$A(\omega_{d1})=\frac{A}{1.5}=\frac{2A}{3}$
When driven frequency is $10\%$ below $\omega_{0}$
$A(\omega_{d1})=\frac{F_{d}}{m\sqrt {(0.9^{2}\omega_{0}^{2}-\omega_{0}^{2})^{2}+\omega_{0}^{2}\times0.9^{2}\omega_{0}^{2}/25}}$
$A(\omega_{d1})= \frac{5F_{d}}{m\omega_{0}^{2}\sqrt {0.9+0.81}}$
$A(\omega_{d1})=\frac{A}{1.3}$
