= Solution
For a reversible <adiabatic process>, an <ideal gas> obeys $TV^{\gamma-1}=\text{constant}$. A <monatomic gas> has <heat capacity ratio> $\gamma=5/3$. Taking the stated expansion factor to mean $V_f=9V_i$,
$$
\boxed{T_f=320\left(\frac19\right)^{2/3}\ \mathrm K
\simeq74.0\ \mathrm K.}
$$
This assumes the usual quasistatic <isentropic process>. Heat insulation alone does not determine the final <temperature>: an ideal-gas <free expansion> is also adiabatic but leaves the temperature unchanged. The numerical answer uses nine times the original volume, the conventional intended meaning of the source's “nine times larger”.
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