Answer
$120$ decibels
Work Step by Step
Substitute $I$ with $10^{12}I_{0}$ in the given formula:
$D=10\displaystyle \log\frac{10^{12}I_{0}}{I_{0}}=10\log 10^{12}$
$... $Basic Logarithmic Properties: $\log_{b}b^{x}=x$...
$D=10\cdot 12=120$ decibels
