Answer
$P_{av,2} = 0.100~W$
Work Step by Step
$P_{av,1} = \frac{1}{2}\sqrt{\mu~F_T}~\omega_1^2~A^2$
We can find an expression for $\omega$.
$\omega = 2\pi~f$
$\omega = 2\pi~(\frac{v}{\lambda})$
$\omega = 2\pi~\sqrt{\frac{F_T}{\mu}}(\frac{1}{\lambda})$
We can find an expression for $\omega_1$.
$\omega_1 = 2\pi~\sqrt{\frac{F_T}{\mu}}(\frac{1}{\lambda_1})$
We can find an expression for $\omega_2$ when $\lambda_2=2\lambda_1$.
$\omega_2 = 2\pi~\sqrt{\frac{F_T}{\mu}}(\frac{1}{\lambda_2})$
$\omega_2 = 2\pi~\sqrt{\frac{F_T}{\mu}}(\frac{1}{2\lambda_1})$
$\omega_2 = (\frac{1}{2})~2\pi~\sqrt{\frac{F_T}{\mu}}(\frac{1}{\lambda_1})$
$\omega_2 = \frac{\omega_1}{2}$
We can find the new average power.
$P_{av,2} = \frac{1}{2}\sqrt{\mu~F_T}~\omega_2^2~A^2$
$P_{av,2} = \frac{1}{2}\sqrt{\mu~F_T}~(\frac{\omega_1}{2})^2~A^2$
$P_{av,2} = (\frac{1}{4})~\frac{1}
{2}\sqrt{\mu~F_T}~\omega_1^2~A^2$
$P_{av,2} = (\frac{1}{4})~P_{av,1}$
$P_{av,2} = (\frac{1}{4})~(0.400~W)$
$P_{av,2} = 0.100~W$
