The answers are all here. If $f(t) = t^n$ then $\mathcal{L}(f)(s) = \frac{n!}{s^{n+1}}$. So the Laplace transform of $6t^3$ is $\frac{36}{s^4}$. Also, the Laplace transform converts convolution products to regular products. So $\mathcal{L}((f*f)(t)) = (\mathcal{L}(f)(s))^2$. Therefore $(\mathcal{L}(f)(s))^2 = \frac{36}{s^4}$. Take the square root of both sides to get $\mathcal{L}(f)(s)$, then look up the inverse Laplace transform of that to find f(t).