The empirical formula for butane is
\[C_4H_{10}\]
, so that 1 mol of butane has a mass of \[12 \times 4 +1 \times 10=58 \; g = 0.058 \; kg\]
In 15 kg butane there are
\[\frac{15}{0.058} = 258.6 \: mol\]
A 15 kg butane bottle contains 15 kg butane, so \[15 \times 1.67 \: litres = 25 \times 10^{-3} m^3\]
Take the temperature as
\[25^{\circ} C \equiv 298^{\circ} K\]
Now use the ideal gas equation
\[pV=nRT\]
\[p= \frac{nRT}{V} = \frac{258.6 \times 8.314 \times 298}{25 \times 10^{-3}} = 2.563 \times 10^7 Pa\]
Atmospheric pressure is about
\[10^5 \; Pa\]
so this is equivalent to about \[\frac{2.568 \times 10^7}{10^5} = 256.8 \]
times atmospheric pressure.