I will choose to solve a somewhat easier problem: the cardinality of [math]\ \text{GL}_3(\Bbb F_q)[/math].
Since these are 3x3 matrices with non-zero determinant, their columns must be linearly independent. We can choose any non-zero vector in [math]\Bbb (F_q)^3[/math] as our first column, there are [math]q^3 - 1[/math] of these. For our second column, we can pick any non-zero vector that is not a scalar multiple of the first. This gives: [math]q^3 -1 - (q - 1) = q^3 - q[/math] choices for the second column, as there are [math]q - 1[/math] non-zero scalars to choose from. Finally, there are [math]q^2 - 1[/math] non-zero linear combinations of the first two chosen vectors, which we must exclude from our choice for the 3rd column, leaving: [math]q^3 - 1 - (q^2 - 1) = q^3 - q^2[/math] choices for column 3. Thus:
[math]|\text{GL}_3(\Bbb F_q)| = (q^3-1)(q^3-q)(q^3-q^2)[/math]
Now we have the short exact sequence:
[math]0 \to \text{SL}_3(\Bbb F_q) \to \text{GL}_3(\Bbb F_q) \xrightarrow{det} (\Bbb F_q)^{\ast} \to 1[/math]
Which tells us that:
[math]|\text{SL}_3(\Bbb F_q)| = \frac{|\text{GL}_3(\Bbb F_q)|}{|(\Bbb F_q)^{\ast}|} = \frac{(q^3-1)(q^3-q)(q^3-q^2)}{q-1} = q^3(q^2+q+1)(q-1)^2(q+1)[/math]CLT Note: I just left my answer as $q^2(q^3-1)(q^3-q)$, but it doesn't really matter.
