Heisenberg's uncertainty principle is
$$\Delta x \Delta p \geq \hbar/2.$$
Since the well is of width $L$, you have a measure for the uncertainty on the position $\Delta x$. Then assume the lowest possible value for $\Delta p$, i.e. the one for which the above inequality becomes an equality. Lastly, use $E = \dfrac{p^2}{2m}$ to find an expression for $E$.
A useful question to look at as well might be this one.
Disclaimer: In this answer, we will just derive a rough semiclassical estimate for the threshold between the existence of zero and one bound state. Of course, one should keep in mind that the semiclassical WKB method is not reliable$^1$ for predicting the ground state. We leave it to others to perform a full numerical analysis of the problem using Airy Functions.
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$\uparrow$ Fig.1: Potential $V(x)$ as a function of position $x$ in OP's example.
First let us include the metaplectic correction/Maslov index. The turning point at an infinitely hard wall and an inclined potential wall have Maslov index $2$ and $1$, respectively, cf. e.g. this Phys.SE post. In total $3$. We should then adjust the BohrSommerfeld quantization rule with a fraction $\frac{3}{4}$.
$$ \int_{x_}^{x_+} \! \frac{dx}{\pi} k(x)~\simeq~n+\frac{3}{4},\qquad n~\in~\mathbb{N}_0,\tag{1} $$
where
$$ k(x)~:=~\frac{\sqrt{2m(EV(x))}}{\hbar}, \qquad
V(x)~:=~V_0 \frac{Lx}{L}. \tag{2} $$
At the threshold, we can assume $n=0$ and $E=0$. The limiting values of the turning points are $x_=0$ and $x_+=L$. Straightforward algebra yields that the
threshold between the existence of zero and one bound state is
$$V_0~\simeq~\frac{81}{128} \frac{\pi^2\hbar^2}{mL^2} \tag{3} .$$
$^1$ For comparison, the WKB approximation for the threshold of the corresponding square well problem yields
$$V_0~\simeq~\frac{\pi^2\hbar^2}{2m L^2} \tag{4} ,$$
while the exact quantum mechanical result is
$$V_0~=~\frac{\pi^2\hbar^2}{8m L^2} \tag{5} ,$$
cf. e.g. Alonso & Finn, Quantum and Statistical Physics, Vol 3, p. 7778. Not impressive!
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$\uparrow$ Fig.2: Corresponding square well potential as a function of position $x$. Each of the 2 infinitely hard walls has Maslow index 2.
Best Answer
The estimate only applies, when the particle is really pressed into the well ($V_0 = \infty$). However making the potential close to zero, the particle's wavefunction expands outside due to tunneling. In your estimation, $a$ should not be the diameter of the well, but rather the characteristic diameter of the wavefunction. This goes arbitrarily large (when potential goes arbitrarily small), thus there are no lower limits to the kinetic energy.