Answer
Larger by: $\log{20}\approx 1.3$
Work Step by Step
We know that $M=\log\frac{I}{S}$ and thus $10^M=\frac{I}{S}$
Hence, here if $I_1,I_2$ are the intensities, where $I_1=20I_2$, we have (by the product rule for logarithms):
$M_1=\log\frac{20I_2}{10^{-4}}=\log\frac{I_2}{10^{-4}}+\log{20}=M_2+\log{20}\approx 1.3$ (magnitude larger)
