Answer
The volume occupied outside the tire is $0.0614~m^3$
Work Step by Step
We can convert the gauge pressure of 36.0 psi to units of $Pa$:
$P_g = 36.0~psi \times \frac{6895~Pa}{1~psi} = 248,000~Pa$
The absolute pressure initially:
$P_1 = P_{atm}+P_g$
$P_1 = (1.01\times 10^5~Pa)+(2.48\times 10^5~Pa)$
$P_1 = 3.49\times 10^5~Pa$
We can find an expression for the original volume:
$P_1~V_1 = nRT$
$V_1 = \frac{nRT}{P_1}$
We can find an expression for the final volume:
$P_2~V_2 = nRT$
$V_2 = \frac{nRT}{P_2}$
We can divide $V_2$ by $V_1$:
$\frac{V_2}{V_1} = \frac{\frac{nRT}{P_2}}{\frac{nRT}{P_1}}$
$V_2 = \frac{P_1}{P_2}~V_1$
$V_2 = \left(\frac{3.49\times 10^5~Pa}{1.01\times 10^5~Pa}\right)~(0.0250~m^3)$
$V_2 = 0.0864~m^3$
Since the volume inside the tube is still $0.0250~m^3$ the volume occupied outside the tire is $0.0864~m^3-0.0250~m^3$ which is $0.0614~m^3$
