$$L[J_{0}(t)] = \frac{1}{\sqrt{s^2 + 1}}$$ -> given

By change of scale property

$L[J_{0}(4t)] = \frac{1}{4}\frac{1}{\sqrt{\frac{s^2}{4^2} +1}} = \frac{1}{4}*\frac{4}{\sqrt{s^2 + 16}} \\ L[tJ_{0}4t] = -1\frac{d}{ds}\frac{1}{\sqrt{s^2 + 16}}$

By multiplication property

$\therefore L[tJ_{0}4t] = -1\frac{d}{ds}\frac{1}{(s^2 + 16)^{\frac{1}{2}}}$

$\therefore L[tJ_{0}4t] = -1\frac{d}{ds}[s^2 + 16]^{\frac{-1}{2}}$

$\therefore L[tJ_{0}4t] = -1\frac{-1}{2}(s^2 + 16)^{\frac{-3}{2}}\frac{d}{ds}(s^2 + 16)$

$\therefore L[tJ_{0}4t] = \frac{1}{2}\frac{1}{(s^2 + 16)^{\frac{3}{2}}}*2s$

$\therefore L[tJ_{0}4t] = \frac{s}{(s^2 + 16)^{\frac{3}{2}}}$

Now by definition of Laplace transform

$\int\limits_{0}^{\infty}e^{-6t}tJ_{0}(4t)dt = \frac{s}{(s^2 + 16)^{\frac{3}{2}}}$

Substituting s = 6

$\therefore \int\limits_{0}^{\infty}e^{-6t}tJ_{0}(4t)dt = \frac{6}{(6^2 + 16)^{\frac{3}{2}}}$

$\therefore \int\limits_{0}^{\infty}e^{-6t}tJ_{0}(4t)dt = \frac{6}{52\sqrt{52}}$

$\therefore \int\limits_{0}^{\infty}e^{-6t}tJ_{0}(4t)dt = \frac{3}{26\sqrt{52}} \neq \frac{3}{500}$