Answer
When the sound level increases by 1 dB, the sound intensity increases by a factor of 1.26
Work Step by Step
Let $\beta_2 = 61~dB$ when the intensity is $I_2$. Let $\beta_1 = 60~dB$ when the intensity is $I_1$. Therefore;
$\beta_2-\beta_1 = 10~log(\frac{I_2}{I_0})-10~log(\frac{I_1}{I_0})$
$61~dB-60~dB = 10~[log(\frac{I_2}{I_0})-log(\frac{I_1}{I_0})]$
$1~dB = 10~log\frac{(\frac{I_2}{I_0})}{(\frac{I_1}{I_0})}$
$0.1 = log\frac{I_2}{I_1}$
$10^{0.1} = \frac{I_2}{I_1}$
$I_2 = 10^{0.1}~I_1$
$I_2 = 1.26~I_1$
When the sound level increases by 1 dB, the sound intensity increases by a factor of 1.26.
