Answer
$$k = \frac{1}{2}$$
Work Step by Step
$$\eqalign{
& {\text{Let }}f\left( x \right) = \frac{1}{{{k^2} + {x^2}}},{\text{ }}\left[ { - k,k} \right]{\text{ and }}{f_{{\text{ave}}}} = \pi \cr
& {\text{Using the definition }}{f_{{\text{ave}}}} = \frac{1}{{b - a}}\int_a^b {f\left( x \right)dx} \cr
& \pi = \frac{1}{{k - \left( { - k} \right)}}\int_{ - k}^k {\frac{1}{{{k^2} + {x^2}}}dx} \cr
& \int_{ - k}^k {\frac{1}{{{k^2} + {x^2}}}dx} = \pi \cr
& {\text{Integrating}} \cr
& \frac{1}{k}\left( {{{\tan }^{ - 1}}\frac{x}{k}} \right)_{ - k}^k = \pi \cr
& {\text{evaluating}} \cr
& \frac{1}{k}\left( {{{\tan }^{ - 1}}\frac{k}{k} - {{\tan }^{ - 1}}\frac{{ - k}}{k}} \right) = \pi \cr
& \frac{1}{k}\left( {{{\tan }^{ - 1}}1 - {{\tan }^{ - 1}}\left( { - 1} \right)} \right) = \pi \cr
& \frac{1}{k}\left( {\frac{\pi }{4} + \frac{\pi }{4}} \right) = \pi \cr
& {\text{Solve for }}k \cr
& \frac{1}{k}\left( {\frac{\pi }{2}} \right) = \pi \cr
& k = \frac{\pi }{{2\pi }} \cr
& k = \frac{1}{2} \cr} $$
