Flat basketball

First of all, it's not a mechanical problem. It's a thermodynamics problem!

From the ideal gas law we have $\frac{PV}{T} = \text{constant,}$ where $P$ is absolute pressure (gauge pressure $+$ atmospheric pressure) and $T$ is absolute temperature ( $\text{t } ^{\circ}\text{C} + 273.15$ ). Since the basketball's volume is assumed constant, $V_1 = V_2$ , we obtain $\begin{aligned} \frac{P_1V_1}{T_1} &= \frac{P_2V_2}{T_2}\\ \frac{60,000+101,325}{20+273.15} &= \frac{P_{gauge}+101,325}{10+273.15}\\ P_{gauge} &\approx \boxed{54,494.027 \text{ Pa}} \end{aligned}$

Pressure is directly proportional to temperature, so $\frac{\text{Pressure}}{\text{Temperature}}$ is constant. Therefore (correcting for gauge pressure and converting to Kelvin): $\frac{60000+101325}{20+273.15}=\frac{P_{\text{new}}+101325}{10+273.15}.$ Hence $P_{\text{new}}=54496\approx 54500.$