3 5 3 5 3 5

Let,

$x = \sqrt{3 \sqrt{5 \sqrt{3 \sqrt{5 .....}}}}$

Now squaring both sides, we get

$x^{2} = 3 \sqrt{5 \sqrt{3 \sqrt{5 .....}}}$

Again squaring both sides, we get

$x^{4} = 9 \times 5 \sqrt{3 \sqrt{5 \sqrt{3 \sqrt{5 .....}}}}$

$\Rightarrow x^{4} = 45x$

$\Rightarrow x^{3} = 45$

$\Rightarrow x = \sqrt[3]{45}$

$\begin{aligned} A&=&\ \ {\sqrt{3\sqrt{5\sqrt{3\sqrt{5\cdots}}}}} \\ \,&=&\ \ \sqrt{\sqrt{3^2\cdot5\sqrt{\sqrt{3^2\cdot5\cdots}}}}\\ \,&=&\ \ \sqrt[4]{45\sqrt[4]{45\sqrt[4]{45\sqrt[4]{45\cdots}}}}\\ \,&=&\ \ \sqrt[4]{45}\cdot \sqrt[16]{45}\cdot \sqrt[64]{45}\cdots\\ \,&=&\ \ 45^{\frac{1}{4}+\frac{1}{16}+\frac{1}{64}+\dots}\\ \end{aligned}$
$\frac{1}{4}+\frac{1}{16}+\frac{1}{64}+\dots = \frac{1}{4}\cdot \frac{1}{1-\frac{1}{4}}=\frac{1}{3}$
$A=\sqrt[3]{45}$

$\text{Let}\, y = \sqrt{\color{#3D99F6}3\sqrt{\color{#20A900}5\sqrt{\color{#3D99F6}3\sqrt{\color{#20A900}5 \cdots}}}} \\ \text{therefore}\, y = \sqrt{\color{#3D99F6}3\sqrt{\color{#20A900}5y}} \\ \frac{y^2}{\color{#3D99F6}3} = \sqrt{\color{#20A900}5y}\\ \frac{y^4}{45}=y\\ y^4 - 45y=0\\ y(y^3-45)=0 \\ \text{either}\, y=0\quad \text{or}\, y= \sqrt[3]{45}\\ \text{since}\, y \ne 0 \quad \text{we have}\, y= \sqrt[3]{45}$

$\sqrt{3\sqrt{5\sqrt{3\sqrt{5\cdots}}}} = x = (\frac{x^2}{3})^2 \times\dfrac{1}{5} = \frac{x^4}{45}\implies\ 45x = x^4 \therefore x = 45^{\frac{1}{3}}$