Finding change in activation energy from change in rate constant in catalysis
It appears that you are correct in your thought process.
Determining by how much the activation energy of the reaction is reduced requires you to basically find

$$Δ(\%) = \frac{E_\mathrm{a1} - E_\mathrm{a2}}{E_\mathrm{a1}} × 100\% = \left(1 - \frac{E_\mathrm{a2}}{E_\mathrm{a1}}\right) × 100\%$$

There is a system of three independent equations

$$
\begin{cases}
k_1 = A\mathrm e^{E_\mathrm{a1}/(RT)}\\
k_2 = A\mathrm e^{E_\mathrm{a2}/(RT)}\\
\frac{k_2}{k_1} = x
\end{cases}
$$

with five unknowns $k_1$, $k_2$, $E_\mathrm{a1}$, $E_\mathrm{a2}$, $A$, which requires either one more equation, or to know one more parameter of the system in order to find $E_\mathrm{a2}/E_\mathrm{a1}.$
The textbook provides you with activation energy in the previous illustration right next to your problem [1, p. 841], so I assume you are supposed to use $E_\mathrm{a1} = \pu{76 kJ mol−1}:$


  


At this point the problem can indeed be solved to a certain numerical value:

$$E_\mathrm{a1} - E_\mathrm{a2} = RT\ln x = \pu{8.3145 J mol-1 K-1} × \pu{298 K} × \ln{2000} \approx \pu{18.8 kJ mol-1}$$

$$Δ(\%) = \frac{\pu{76 kJ mol-1} - \pu{18.8 kJ mol-1}}{\pu{76 kJ mol-1}} × 100\% \approx 25\%$$

The question whether it's a textbook flaw, or authors assumed the reader would read the material more carefully still remains open.

References


Atkins, P. W.; De Paula, J. Physical Chemistry: Thermodynamics, Structure, and Change, 10th ed.; W.H. Freeman: New York, 2014. ISBN 978-1-4641-2452-5.