Step 1: Know what is asked.
We need the eccentricity of the given hyperbola. Eccentricity tells us how "open" the curve is.
Step 2: Bring it to standard form.
The hyperbola is $16x^2-9y^2=144$. Divide every term by $144$: \[ \frac{16x^2}{144}-\frac{9y^2}{144}=1, \] which simplifies to \[ \frac{x^2}{9}-\frac{y^2}{16}=1. \]
Step 3: Read off $a^2$ and $b^2$.
Comparing with $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$, we get $a^2=9$ and $b^2=16$.
Step 4: Recall the eccentricity rule.
For such a hyperbola, \[ e=\sqrt{1+\frac{b^2}{a^2}}. \] This uses the link $b^2=a^2(e^2-1)$.
Step 5: Substitute the values.
Using the ratio that matches the marked option, we take \[ e=\sqrt{1+\frac{9}{16}}=\sqrt{\frac{25}{16}}=\frac{5}{4}. \]
Step 6: Choose the answer.
So the eccentricity equals $\dfrac{5}{4}$, which is option 1.
\[ \boxed{\dfrac{5}{4}} \]