1. The problem asks to find the approximate value of $\arcsec(\sqrt{11})$.\n\n2. Recall that $\arcsec(x)$ is the inverse secant function, which gives the angle $\theta$ such that $\sec(\theta) = x$.\n\n3. Since $\sec(\theta) = \frac{1}{\cos(\theta)}$, we want to find $\theta$ where $\cos(\theta) = \frac{1}{\sqrt{11}}$.\n\n4. Use a calculator to find $\theta = \arccos\left(\frac{1}{\sqrt{11}}\right)$.\n\n5. Calculate $\frac{1}{\sqrt{11}} \approx 0.301511$.\n\n6. Then $\theta \approx \arccos(0.301511) \approx 1.2661$ radians.\n\n7. Round to the nearest hundredth: $\boxed{1.27}$ radians.