Answer
We use $$\beta=10 \log \left(I / I_{0}\right)$$ with $$I_{0}=1 \times 10^{-12} \mathrm{W} / \mathrm{m}^{2}$$
and Eq. $17-27$ with
$$
\omega=2 \pi f=2 \pi(260 \mathrm{Hz})
$$
$$v=343 \mathrm{m} / \mathrm{s}$$ and $$\rho=1.21 \mathrm{kg} / \mathrm{m}^{3} .$$
$$I=I_{0}\left(10^{8.5}\right)=\frac{1}{2} \rho v(2 \pi f)^{2} s_{m}^{2} \quad \Rightarrow \quad s_{m}=7.6 \times 10^{-7} \mathrm{m}=0.76 \mu \mathrm{m}$$
Work Step by Step
We use $$\beta=10 \log \left(I / I_{0}\right)$$ with $$I_{0}=1 \times 10^{-12} \mathrm{W} / \mathrm{m}^{2}$$
and Eq. $17-27$ with
$$
\omega=2 \pi f=2 \pi(260 \mathrm{Hz})
$$
$$v=343 \mathrm{m} / \mathrm{s}$$ and $$\rho=1.21 \mathrm{kg} / \mathrm{m}^{3} .$$
$$I=I_{0}\left(10^{8.5}\right)=\frac{1}{2} \rho v(2 \pi f)^{2} s_{m}^{2} \quad \Rightarrow \quad s_{m}=7.6 \times 10^{-7} \mathrm{m}=0.76 \mu \mathrm{m}$$
