 Solution

Write $E_{\ast }=({\hslash }^2{a}^4/(2m){)}^{1/5}$. Then $E_0^{\mathrm{WKB}}=1.811694E_{\ast }$ and $E_{\mathrm{var}}=1.876493E_{\ast }$. The variational method guarantees $E_0\le E_{\mathrm{var}}$, 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.