Answer
The intensity is increased by a multiple of $~~1000$
Work Step by Step
Let $I_1$ be the initial sound intensity.
Then: $I_1 = 10^{\frac{\beta_1}{10}}~\cdot I_0$
We can find an expression for the sound intensity $I_2$ after the sound level increase:
$\beta_1+30~dB = (10~dB)~log(\frac{I_2}{I_0})$
$\frac{\beta_1}{10}+3 = log(\frac{I_2}{I_0})$
$10^{\frac{\beta_1}{10}+3} = \frac{I_2}{I_0}$
$10^{\frac{\beta_1}{10}}~\cdot 10^3 = \frac{I_2}{I_0}$
$I_2 = (1000)~(10^{\frac{\beta_1}{10}}~I_0)$
$I_2 = 1000 ~I_1$
The intensity is increased by a multiple of $~~1000$
