Given,

$f\left(x\right)=\frac{2}{\sqrt{3}}{\int }_0^{\sqrt{3}}f\left(\frac{{\lambda }^{2}x}{3}\right)d\lambda$

Now let $\frac{{\lambda }^{2}x}{3}=t$

$\Rightarrow \frac{2\lambda x}{3} d\lambda =dt$

$\Rightarrow d\lambda =\frac{3}{2}·\frac{1\sqrt{x}}{\mathrm{x}·\sqrt{3}\sqrt{t}}dt$

$\Rightarrow d\lambda =\frac{\sqrt{3}}{2}·\frac{1}{\sqrt{x}}·\frac{dt}{\sqrt{t}}$

So, $f\left(x\right)=\frac{1}{\sqrt{x}}{\int }_0^x\frac{f\left(t\right)}{\sqrt{t}}dt$

$\sqrt{x}f\left(x\right)={\int }_0^x\frac{f\left(t\right)}{\sqrt{t}}dt$

Now differentiating both side we get,

$\Rightarrow \sqrt{x}·f^{\text{'}}\left(x\right)+\frac{f\left(x\right)}{2\sqrt{x}}=\frac{f\left(x\right)}{\sqrt{x}}$

$\Rightarrow \sqrt{x}·f^{\text{'}}\left(x\right)=\frac{f\left(x\right)}{2\sqrt{x}}$

$\Rightarrow \frac{dy}{y}=\frac{dx}{2x}$

Now integrating both side we get,

$\Rightarrow \int \frac{dy}{y}=\int \frac{dx}{2x}$

$\Rightarrow \ln y=\frac{1}{2}\ln x+\ln c$

Given $f\left(1\right)=\sqrt{3}$, so $\Rightarrow \ln \sqrt{3}=0+\ln c$

$\Rightarrow c=\sqrt{3}$

So, $f\left(x\right)=\sqrt{3x}$

Now given $f\left(x\right)$ is passing through $\left(\alpha ,6\right)$,

So $f\left(\alpha \right)=6\Rightarrow 36=3\alpha$

$\Rightarrow \alpha =12$