I do not know a reference but the following argument gives the best constant. Consider the interval $[0,a]$ and $G(t,s)$ the Green function of $I-D^2$ with zero boundary conditions at $0,a$. If $u \in H^2 \cap H^1_0$, then
$$u(t)=\int_0^a G(t,s)\left (u(s)-u''(s)\right )ds=\int_0^a \left(G(t,s)u(s)+G_s(t,s)u'(s) \right )ds
$$ so that
$$|u(t)| \le \left (\int_0^a G(t,s)^2ds \right )^{1/2}\|u\|_2+\left (\int_0^a G_s(t,s)^2ds \right )^{1/2}\|u'\|_2 \le M_t \|u\|_{H^1}
$$
with $M_t^2=\int_0^a \left(G(t,s)^2+G_s(t,s)^2\right )ds$. This inequality extends to $H^1_0$, by density. Then the best constant $C$ satisfies $C \leq \sup_t M_t$. On the other hand, if $u(\cdot)=G(t, \cdot)$ then
$$G(t,t)=\int_0^a \left(G(t,s)^2+G_s(t,s)^2\right )ds=M_t \|u\|_{H^1}
$$
and hence $C=\sup_t M_t$. At this point one has to compute patiently $G(t,s)$ and $M_t$ and maximize it over $[0,a]$. The maximum is in $a/2$ (as expected) and I got (if no mistake occurred)
$$C=\left(\frac{1}{2} \coth a-\frac{1}{4 \sinh a}\right )^{1/2}
$$ which tends to $1/\sqrt 2$ as $a \to \infty$, the best constant on the line.