= Solution
Assume three-dimensional isotropic gapless excitations, zero <chemical potential> and one independent branch. Momentum-state counting gives
$$
\frac{g(\epsilon)}V=\frac{1}{2\pi^2\hbar^3n}\alpha^{-3/n}\epsilon^{3/n-1}.
$$
Substituting $x=\epsilon/(k_BT)$ into the Bose <energy> integral shows $U/V\propto T^{1+3/n}$ and $C/V\propto T^{3/n}$. The observed exponent therefore gives \b[$n=2$].
For the <quadratic-boson heat-capacity coefficient>, the required Bose integral is
$$
I_{3/2}=\int_0^\infty\frac{x^{3/2}\,dx}{e^x-1}
=\Gamma(5/2)\zeta(5/2)=1.78329.
$$
The equality follows by expanding $(e^x-1)^{-1}=\sum_{r\ge1}e^{-rx}$ and integrating term by term. Hence
$$
\frac UV=\frac{I_{3/2}(k_BT)^{5/2}}{4\pi^2\hbar^3\alpha^{3/2}},\qquad
\frac CV=\frac{5I_{3/2}k_B^{5/2}}{8\pi^2\hbar^3\alpha^{3/2}}T^{3/2}.
$$
The supplied units make its $C$ a <heat capacity> per <volume>. Equating its coefficient to $220\,\mathrm{J\,K^{-5/2}\,m^{-3}}$ gives
$$
\boxed{\alpha=4.58\times10^{27}\,\mathrm{kg^{-1}}.}
$$
Equivalently $\epsilon=p^2/(2m_*)$ has $m_*=1.09\times10^{-28}\,\mathrm{kg}$. For $q_b$ identical branches, the measured coefficient fixes $q_b/\alpha^{3/2}$, so instead $\alpha=(4.58\times10^{27}\,\mathrm{kg^{-1}})q_b^{2/3}$. The one-branch convention is necessary for the quoted numerical value. The supplied integral with $x^4$ is not the moment arising for the exponent determined here.
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