Since $PM= dRT => \large\frac{T}{M} =\frac{P}{dR} =\frac{10^5}{45 \times 8.3}$

now, $V_{rms} =\sqrt { \large\frac{3RT}{M} =\sqrt {\frac 3P}{d}} =\sqrt {\large\frac{3 \times 10^5 \times 3}{25}}$

$\qquad = 3 \times 2 \times \sqrt {10^3}$

Hence C is the correct answer.