Length of perpendicular from origin to the line $x\sqrt{5}+2y=3\sqrt{5}$ is

$OL=\frac{3\sqrt{5}}{\sqrt{{\left(\sqrt{5}\right)}^2+2^2}}=\frac{3\sqrt{5}}{\sqrt{9}}=\sqrt{5}$

Radius of the given circle $=\sqrt{10}=OQ=OP$

$PQ=2QL$

$=2\sqrt{O{Q}^2-O{L}^2}$

$=2\sqrt{10-5}=2\sqrt{5}$

Thus, area of $\mathrm{\Delta }OPQ=\frac{1}{2}\times PQ\times OL$

$=\frac{1}{2}\times 2\sqrt{5}\times \sqrt{5}=5$ sq units

Hence, option (c) is correct.