$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$
Hyperbola passes through $(3 \sqrt{5}, 1)$
$\therefore \quad \frac{(3 \sqrt{5})^{2}}{\mathrm{a}^{2}}-\frac{1}{\mathrm{~b}^{2}}=1$...(i)
$\frac{45}{\mathrm{a}^{2}}-\frac{1}{\mathrm{~b}^{2}}=1$
Now length of latus rectum $=\frac{2 \mathrm{~b}^{2}}{\mathrm{a}}$
$\Rightarrow \frac{4}{3}=\frac{2 \mathrm{~b}^{2}}{\mathrm{a}}$
$\Rightarrow \quad \frac{2}{3}=\frac{\mathrm{b}^{2}}{\mathrm{a}} \Rightarrow \mathrm{a}=\frac{3 \mathrm{~b}^{2}}{2}$...(ii)
Putting the value of ' $a$ ' from equation (ii) in equation (i),
$\Rightarrow \frac{45 \times 4}{9 b^{4}}-\frac{1}{b^{2}}=1$
$\Rightarrow \frac{20}{b^{4}}-\frac{1}{b^{2}}=1$
$20-b^{2}=b^{4}$
$b^{4}+b^{2}-20=0$
$b^{4}+5 b^{2}-4 b^{2}-20=0$
$b^{2}\left(b^{2}+5\right)-4\left(b^{2}+5\right)=0$
$\left(b^{2}-4\right)\left(b^{2}+5\right)=0$
$b^{2}=4, b^{2}=-5$
$\therefore \quad b^{2}=4 \Rightarrow b=2$
Nowlength of conjugate axis $=2 \mathrm{~b}=2(2)=4$ $\therefore$ Option (c) is correct.