= Solution
Write $E_*=(\hbar^2a^4/(2m))^{1/5}$. Then \b[$E_0^{\rm WKB}=1.811694E_*$ and $E_{\rm var}=1.876493E_*$]. The <variational method> guarantees $E_0\le E_{\rm var}$, since the trial satisfies the wall condition and has finite <energy>. WKB gives no comparable upper-bound guarantee and is least controlled for the lowest state. The two estimates differ by about $3.6\%$ of the WKB value; their closeness is encouraging but does not prove that the exact <energy> lies between them.
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