This is a good question and it is not entirely trivial, because there is no such example using a simple cone construction. More generally, if the resolution graph of the (normal) singularity is a chain of rational curves, then if the singularity is log canonical, then it is necessarily log terminal. (I won't include the calculation of this, but it is very similar to the calculation below, relatively straightforward using adjunction.)

So, to get an example like that one needs a singularity with a slightly more complicated resolution graph. The simplest graph that is not a chain is a triple fork. So, let's suppose the exceptional curves are $E_0,E_1,E_2,E_3$ such that $E_0\cdot E_i=1$ for all $i\neq 0$ and $E_i\cdot E_j=0$ for all $i\neq 0, j\neq 0$. Furthermore let $E_i^2=-n_i$ with $n_i\in\mathbb N$ for all $i$. By Artin's criterion this is a rational singularity for any choices where $n_i\geq 2$.

Writing down adjunction for each curve $E_i$ gives the equations:

$$-n_0(a_0+1) +a_1 +a_2+a_3 =-2$$
and

$$ -n_j(a_j+1) + a_0 =-2 $$
for all $j\neq 0$.

These imply that **if** $a_0\geq -1$, then
$$ a_j + 1 = \frac{a_0+2}{n_j} \geq \frac{-1\ \ \ }{n_j} > 0,$$
so
$$ a_j> -1 $$

and then solving for $a_0$ gives
$$ a_0 = -2 + \frac{n_0-1}{n_0-(\frac 1{n_1}+\frac 1{n_2}+\frac 1{n_3})}$$
so this singularity is log canonical, i.e., $a_0\geq -1$ iff
$$
\frac 1{n_1}+\frac 1{n_2}+\frac 1{n_3}\geq 1
$$
and it is log terminal iff there is strict inequality here.

In particular it is log canonical, but not log terminal if there is equality which happens for example if $n_1=2, n_2=3, n_3=6$ and $n_0\geq 2$ arbitrary.

Finally, you might ask whether such a singularity exists, but that's easy. By blowing up appropriate points you can certainly create a configuration as above. Then you check that the intersection matrix is negative definite and use Artin's criterion to contract this configuration. The resulting singularity will be your example.