= Solution
Within the prescribed uniform-density approximation, $n=3M/(4\pi m_pR^3)$. Treating the given approximate <pressure> coefficient as one, the <degeneracy pressure> and specified kinetic-energy relation give
$$
E_{\rm kin}=\frac32\frac{h^2}{m_e}n^{5/3}\frac{4\pi R^3}{3}
=\frac{\alpha}{R^2},
\qquad\alpha=\frac32\frac{h^2}{m_e}\left(\frac3{4\pi}\right)^{2/3}
\left(\frac M{m_p}\right)^{5/3}.
$$
For a uniform sphere, $m(r)=M(r/R)^3$, so shell assembly gives
$$
E_{\rm grav}=-\int_0^R\frac{Gm(r)}r\,dm(r)=-\frac{3GM^2}{5R}.
$$
Thus
$$
\boxed{E_{\rm total}(R)=\frac\alpha{R^2}-\frac\beta R,
\qquad\beta=\frac35GM^2.}
$$
At fixed mass, this <energy> has its minimum at $R_*=2\alpha/\beta$, with $E''(R_*)=2\alpha/R_*^4>0$. Expansion beyond this radius raises the <energy>, while strong compression costs positive <energy> growing as $R^{-2}$, faster than the attractive $R^{-1}$ term. This explains radial stability of the model <white dwarf>. The numerical coefficient in $\alpha$ inherits the approximate <pressure> law; a relativistic degeneracy law would change this stability argument.
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